SOLUTIONS MANUAL For An Introduction to Physical Science, 14e James Shipman, Jerry Wilson, Charles H by StudyGuide

Introduction to Physical Sciences 14e

Part 1

Chapter 1

MEASUREMENT Chapter 1 is important because all quantitative knowledge about our physical environment is based on measurement. Some chapter sections have been reorganized and rewritten for clarity. The 1.2 Section, “Scientific Investigation,” introduces the student to the procedures for scientific investigation. Major terms such as experiment, law, hypothesis, theory and scientific method are introduced. The idea that physical science deals with quantitative knowledge should be stressed. It is not enough to know that a car is going “fast”; it is necessary to know how fast. A good understanding of units is of the utmost importance, particularly with the metric-British use in the United States today. The metric SI is introduced and explained. Both the metric and the British systems are used in the book in the early chapters for familiarity. The instructor may decide to do examples primarily in the metric system, but the student should get some practice in converting between the systems. This provides knowledge of the comparative size of similar units in the different systems and makes the student feel comfortable using what may be unfamiliar metric units. The Highlight, “Is Unit Conversion Important? It Sure Is,” illustrates the importance of unit conversion. The general theme of the chapter and the textbook is the students’ position in his or her physical world. Show the students that they know about their environment and themselves through measurements. Measurements are involved in the answers to such questions as, How old are you? How much do you weigh? How tall are you? What is the normal body temperature? How much money do you have? These and many other technical questions are resolved or answered by measurements and quantitative analyses.

DEMONSTRATIONS Have a meter stick, a yardstick, a timer, one or more kilogram masses, a one-liter beaker or a liter soda container, a one-quart container, and a balance or scales available on the instructor’s desk. Demonstrate the comparative units. The meter stick can be compared to the yardstick to show the difference between them, along with the subunits of inches and centimeters. The liter and quart also can be compared. Pass the kilogram mass around the classroom so that students can get some idea of the amount of mass in one kilogram. Mass and weight may be compared on the balance and scales. When discussing Section 1.6, “Derived Units and Conversion Factors,” have class members guess the length of the instructor’s desk in metric and British units. Then have several students

Introduction to Physical Sciences 14e

independently measure the length with the meter stick and yardstick. Compare the measurements in terms of significant figures and units. Compare the averages of the measurements and estimates. Convert the average metric measurement to British units, and vice versa, to practice conversion factors and to see how the measurements compare. Various metric unit demonstrations are available from commercial sources.

ANSWERS TO MATCHING QUESTIONS a. 19 b. 13 c. 21 d. 14 e. 15 f. 8

g. 10

p. 3 q. 20 r. 16

v. 17 w. 5

s. 22 t. 7 u. 23

h. 2

i. 21

j. 1 k. 9

l. 4

m. 18

n. 6 o. 11

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. c

2. b

3. d

4. b

5. b

6. b

7. d

8. c 9. d 10. c

11. d

12. c 13. a

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. biological 7. shorter

2. hypothesis

8. fundamental

3. scientific method 9. time or second

4. sight

5. limitations

6. greater than

10. one-millionth, 10-6 11. liter

12. mass

ANSWERS TO SHORT-ANSWER QUESTIONS 1. An organized body of knowledge about the natural universe by which knowledge is acquired and tested. 2. Physics, chemistry, astronomy, meteorology, and geology. 3. Observations and Measurements. 4. Hypothesis. 5. A law is a concise statement about a fundamental relationship of nature. A theory is a welltested explanation of a broad segment of natural phenomena. 6. That phenomena must be investigated, not speculated. 7. Sight, hearing, touch, taste, and smell. 8. They have limitations and can be deceived. 9. (a) No. (b) Yes. (c) Good luck. (Use a couple of straight edges to determine.) 10. They are the most basic quantities of which we can think. 11. A fixed and reproducible value. 12. A group of standard units and their combinations. 13. km/hour

Introduction to Physical Sciences 14e

14. Yes, officially adopted in 1899. 15.Kilogram, a platinum-iridium cylinder. 16. Mass. Weight varies with gravity. 17. Meter-kilogram-second, International System of Units, centimeter-gram-second. 18. Base 10 easier to use (factors of 10). 19. mega- (M), kilo- (k), centi- (c), milli- (m) 20. Mass of a cubic liter of water. 21. Cubic meter. 22. kg, m, s, and C (electric charge). 23. The compactness of matter. 24. It is given a new name. 25. No. An equation must be equal in magnitude and units. 26. Yes. And it could be confused with “meters” instead of “miles.” 27. To express measured numbers properly. 28. By reading a measurement value from an instrument and rounding according to the general rules. 29. Two.

ANSWERS TO VISUAL CONNECTION a. meter, b. kilogram, c. second, d. mks, e. foot, f. pound, g. second, h. fps

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Scientific laws explain nature, not to regulate society (legal laws), and scientific laws do not compel, just describe. 2. The liter is larger than the quart, and the kilogram is larger than the pound. However, the kilometer itself is shorter than the mile. 3. Intrinsic properties are invariant. Kilogram cylinder and meterstick subject to wear, dirt, and change. 4. A liter, because it is larger than a quart.

Introduction to Physical Sciences 14e

5. (a) No, gold would weigh about 20 times more. (b) Solving for mass from the density equation, this would be 19320 g or 19.32 kg. No playing catch, as weight is about 45 lb. (1 kg = 2.2 lb). 6. 1 m = 3.28 ft. 830 m (3.28 ft/m) = 2.72 x 103 ft; 508 m (3.28 ft/m) = 1.67 x 103 ft. Δ = 1.05 x 103 ft

ANSWERS TO EXERCISES 1. 10,000 cm or 105 cm 2. 2 x l03 Mb 3. 106 mm3 4. 1 m3 = 103 L. 1 m3 = 102 cm x 102 cm x 102 cm = 106 cm3 (1 L/103 cm3) = 103 L = 1000 L. 5. 0.50 L (1 kg/L) = 0.50 kg = 500 g 6. 25 cm  25 cm  35 cm = 22  103 cm3 or mL = 22 L and 12 L (1 kg/L) = 12 kg = 12,000 g 7. (a) 0.55 Ms = 0.55  106 s (b) 2.8 km = 2.8  103 m (c) 12 mg = 12  10–3 g = 1.2  10–6 kg (d) 100 cm = 1.00 m 8. (a) 40 Mb (b) 572.2 mL (c) 540.0 x 102 cm (d) 5.5 kilobucks 9. 6 ft 5 in. = 77 in (2.54 cm/in.) = 196 cm = 1.963 m 10. 6 ft, 7 in. 11. Yes, to two significant figures 12. (a) 55 mi/h (1.61 km/mi) = 89 km/h (90 km/h). (b) 65 mi/h (1.61 km/mi) = 106 km/h (110 km/h) 13. No, 300 L ~ 300 qt (1 gal/4 qt) = 75 gal 14. No. Room would be about 12 m  10 m, which would be about 36 ft.  30 ft. Much too large for a dorm room. 15. See AYK # 6, 1,67 x 103 ft. 16. 890 ft (1 m/3.28 ft) 271 m; 1,900 ft = 579 m 17. cm, km 18. 103 kg (2.2 lb/kg) = 2,200 lb. 103 kg heavier by 200 lb 19.  = m/V = 500 g/63 cm3 = 7.9 g/cm3 (the density of iron) 20. V = m /  = 500 g/ 7.9 g/cm2 = 63 cm3 21. (a) 7,7 (b) 0.0021 (c) 9400 (d) 0.00034

Introduction to Physical Sciences 14e

22. (a) 0.00999 (b) 645 (c) 0.0106 (c) 8430 23. (a) 1.00 (b) 7380 (c) 0.00179 (d) 47.6 24. (a) 3.142 (b) 0.006907 (c) 483.6 (d) 0.02350 25. (3.15 m  1.53 m)/0.78 m = 6.2 m 26. 6.75

(3 sf)

Introduction to Physical Sciences 14e

Chapter 2

MOTION This chapter covers the basics of the description of motion. The concepts of position, speed, velocity, and acceleration are defined and physically interpreted, with applications to falling objects, circular motion, and projectiles. A distinction is made between average values and instantaneous values. Scalar and vector quantities are also discussed. Also, an interesting Highlight on Galileo and the Leaning Tower of Pisa discusses the status of the tower. Problem solving is difficult for most students. The authors have found it successful to assign a take-home quiz on several questions and exercises at the end of the chapter that is handed in at the beginning of class. (It may save time and be instructive to have students exchange and grade papers as you go over the quiz.) This may be followed by an in-class quiz on one of the take-home exercise, for which the numerical values have been changed. The procedure provides students with practice and helps them gain confidence.

DEMONSTRATIONS A linear air track may be used to demonstrate both velocity and acceleration. If an air track is not available, a 2-in.  6-in.  12-ft wooden plank may be substituted. It will be necessary to have a V groove cut into one edge of the plank to hold a steel ball of about 1-in. diameter. The ball will roll fairly freely in the V groove. Also, various free-fall demonstrations are commercially available. (General references to teaching aids are given in the Teaching Aids section.)

ANSWERS TO MATCHING QUESTIONS a. 16

b. 13 c. 1 d. 6

o. 17

p. 11

q. 4

e. 14

f. 2

g. 3

h. 12 i. 5

j. 15 k. 18

l. 8

m. 10 n. 7

r. 9

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d

2. c

3. d

4. d

5. d

7. c

8. d

9. d

10. d

11. c

6. b 12. c

Introduction to Physical Sciences 14e

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. position 2. scalar

3. vector

4. distance

5. speed

6. constant or uniform

7. time, t2 8. gravity

9. m/s2 10. centripetal (center-seeking)

11. 4

12. acceleration

ANSWERS TO SHORT-ANSWER QUESTIONS 1. Mechanics. 2. An origin or reference point. 3. Length per time (length/time). 4. A scalar has magnitude, and a vector has magnitude and direction. 5. Distance is the actual path length and is a scalar. Displacement is the directed, straight-line distance between two points and is a vector. Distance is associated with speed, and displacement is associated with velocity. 6. They both give averages of different quantities. 7. (a) They are equal. (b) The average speed has a finite value, but the average velocity is zero because the displacement is zero. 8. Either the magnitude or direction of the velocity, or both. An example of both is a child going down a wavy slide at a playground. 9. Yes, both (a) and (b) can affect speed and therefore velocity. 10. No. If the velocity and acceleration are both in the negative direction, the object will speed up. 11. Initial speed is zero. Initial acceleration of 9.8 m/s2, which is constant. 12. The object would remain suspended. 13. Yes, in uniform circular motion, velocity changing direction, centripetal acceleration. 14. Center-seeking. Necessary for circular motion. 15. Yes, we are in rotational or circular motion in space. 16. Inwardly toward the Earth's axis of rotation for (a) and (b). 17. g and vx 18. Greater range on the Moon, gravity less (slower vertical motion). 19. Initial velocity, projection angle, and air resistance. 20. No, it will always fall below a horizontal line because of the downward acceleration due to gravity.

Introduction to Physical Sciences 14e

21. Both have the same vertical acceleration. 22. Less than 45o because air resistance reduces the velocity, particularly in the horizontal direction. ANSWERS TO VISUAL CONNECTION a. speed, b. uniform velocity, c. acceleration (change in velocity magnitude), d. acceleration (change in velocity magnitude and direction) ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. More instantaneous. Think of having your speed measured by a radar. This is an instantaneous measurement, and you get a ticket if you exceed the speed limit. 2. (a) The orbital (tangential) acceleration is small and not detected. (b) The apparent motion of the Sun, Moon, and stars. 3. (a) toward the center of the Earth, (b) toward the axis, (c) zero 4. Yes, neglecting air resistance. 5. d  ½ gt 2 , so t  2d / g 

2(11 m)  1.5 s Balloon lands in front of prof. Student gets 9.8 m/s2

an “F” grade. 6. (a) updraft, slow down, reach terminal velocity later. (b) downdraft, speed up, terminal velocity sooner. 7. Escaping air stabilizes chute – prevents rocking. 8. Streamlines. Prevents air blocking. ANSWERS TO EXERCISES 1. 7 m 2. 5 m south of east 3. v = d/t = 100 m/12 s = 8.3 m/s 4. 1.6 m/s 5. t = d/v = 7.86  1010 m/ 3.00  108 m/s = 2.62  l02 s. Speed of light (constant). 6.. t = d/v = 750 mi/(55.0 mi/h) = 13.6 h 7. (a) d = v t = (52 mi/h)(1.5 h) = 78 mi (b) v = d/t = 22 mi/0.50 h = 44 mi/h (c) v = d/t = 100 mi/2.0 h = 50 mi/h 7. v = d/t = 7.86  1010 m/ 2.62  l02 s = 3.00  108 m/s. Speed of light (constant).

Introduction to Physical Sciences 14e

8. (a) d/150 s. (b) d/192 s., (c) d/342 s. Omission. d inadvertently left out. Assuming 100 m, (a) 100 m/150 s = 0.667 m/s. (b) 100 m/192 s = 0.521 m/s. (c) 200 m/ 342 s = 0.585 m/s. 9. (a) v = d/t = 300 km/2.0 h = 150 km/h, east. (b) Same, since constant. 10. (a) v = d/t = 750 m/20.0 s = 37.5 m/s, north. (b) Zero, since displacement is zero. 11. a = (vf – vo )/t = (12 m/s – 0)/6.0 s = 2.0 m/s2 12. (a) a = (vf – vo )/t = (0 – 8.3 m/s)/1200 s = –6.9  10–3 m/s2 (b) v = d/t = (5.0 

m)/(1.2 

s) = 4.2 m/s (Needs to start slowing in plenty of time.)

13. (a) a = (vf – vo )/t = (8.0 m/s – 0)/10 s = 0.08 m/s2 in direction of motion. (b) a = (12 m/s – 0)/15 s = 0.80 m/s2 in direction of motion. 14. (a) (a) 44 ft/s/5.0 s = 8.8 ft/s2, in the direction of motion. (b) 11 ft/s2, (c) -7.3 ft/s2 (b) a = (88 ft /s – 44 ft /s)/4.0 s = 11 ft /s2 in direction of motion. (c) (66 ft /s – 88 ft /s)/3.0 s = –7.3 ft /s2 opposite direction of motion. (d) a = (66 ft /s – 0)/12 s = 5.5 ft /s2 in direction of motion. 15. No, d = ½ gt 2 = ½ (9.8 m/s2) (4.0)2 = 78 m in 4.0 s. 16. v = vo + gt = 0 + (9.8 m/s2)(3.5 s) = 34 m/s 17. d = ½ gt2, t = sq.root [2(2.71 m)/9.80 m/s2] =7.4 s 18. d = ½ gt2. t as in 17. 4.3 s – 2.5 s = 1.8 s. 19. (a) ac = v2/r = (10 m/s)2/ 70 m = 1.4 m/s2 toward center. (b) ac /g = (1.4 m/s2 )/(9.8 m/s2 ) = 0.14 or 14%‚ yes. 20. 90.0 km/h = 25.0 m/s. ac = v2/r = (25.0 m/s)2/500 m = 1.25 m/s2. 21. 0.55 s. Vertical distance is the same. 22. 45o – 37o = 8o, so 45o + 8o = 57o.

Introduction to Physical Sciences 14e

Chapter 3

FORCE AND MOTION This chapter is one of the most important in the textbook because it deals with Newton’s laws of motion and gravitation, as well as the concepts of linear and angular momentum. Also, involving a force, buoyancy and Archimedes’ principle is discussed in this chapter. The material naturally follows that of Chapter 2. With the foundations of kinematics established, the agents that produce motion are considered. This branch of mechanics is known as dynamics. Sufficient time should be spent on this material to be sure that students have a firm understanding of these concepts. Force and net force are discussed in an initial chapter section because of the importance of understanding these concepts. It is suggested that students be required to make complete statements of Newton’s laws and to give examples. When stating Newton’s second law of motion, stress that the force is the unbalanced force acting on the total mass, and that the mass is the total mass being accelerated. Also that the acceleration is in the direction of the unbalanced, or net, force. Acceleration (or deceleration) is evidence of the action of an unbalanced force. The Highlight on automobile airbags includes side airbags and “depowering” features. Also new to the thirteenth edition is the Highlight: Surface Tension, Water Striders, and Soap Bubbles.

DEMONSTRATIONS A linear air track can be used to illustrate Newton’s laws of motion and the concept of linear momentum. An Atwood machine provides an excellent demonstration for illustrating Newton’s second law of motion. Best results can be obtained if the student is led through the demonstration (see the Laboratory Guide) by questions rather than having the instructor merely perform the experiment. An apple may be brought to class to illustrate the idea of one newton of weight. Be sure the apple weighs about 3.6 ounces. Also, a spring balance calibrated in newtons can be displayed supporting a 1-kg mass. Free-fall can be demonstrated with a feather and a coin in a glass tube from which the air can be evacuated. Let your students handle the glass tube for the best results. Newton’s third law of motion can be demonstrated by using a toy rocket that holds water under pressure-equal and opposite forces are demonstrated as the rocket accelerates along a string. Releasing a blown balloon also illustrates the law.

Introduction to Physical Sciences 14e

The law of conservation of angular momentum is demonstrated dramatically using a turntable or rotating stool and two masses (for example, 1 kg each) held in the hands. While rotating, the masses are brought closer to the body (reduced moment of inertia), and the rate of rotation increases. When beginning the demonstration, point out that you can't get started by yourself. You must have an external force or torque, which can be supplied by a student. Students often will ask to try the demonstration. Permission should be granted with caution. The rotation can make a person quite dizzy. (General references to teaching aids are given in the Teaching Aids section.)

ANSWERS TO MATCHING QUESTIONS a. 14 b. 5 c. 11 d. 16 i. 19

j. 10 k. 1

e. 2

l. 12

f. 9

m. 7

g. 15

h. 6

n. 4 o. 17

p. 13 q. 8

r. 18 s. 3

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d

2. d

3. c

4. d

5. a

6. d

7. d

8. d

9. c

10. a 11. c

12. a

13. b

14. d 15. c

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. capable

2. vector

7. kg  m/s

8. static, kinetic or sliding 9. different

3. could

4. force

5. mass 6. inversely 10. universal

11. greater

12. more 13.

net or unbalanced 14. torque

ANSWERS TO SHORT-ANSWER QUESTIONS 1. No. The force may be balanced and the net force zero. 2. They are the same. 3. (a) goes forward (b) goes backward 4. Dishes and glasses remain in place because of inertia. 5. Inertia of hammer head continues motion and tightens on handle. 6. A quick jerk gives inertia for tearing. Larger rolls have more inertia and resistance to motion. 7. If a is zero in F = ma, then F = 0, and an object is at rest or moving with a constant velocity (first law). 8. Yes if the sum of the forces is zero. 9. Yes, by the first law of motion. 10. Zero, since velocity is constant.

Introduction to Physical Sciences 14e

11. Less force to keep moving that to initiate motion. 12. (a) Ten times the force, but also ten times the inertia, or mass; therefore, it falls at the same rate. (b) Similar, except that the acceleration of the rocks would be less—that is, g/6. 13. No, the pad is there only to hold the rocket. The expanding combustion gases exert a force on the rocket, and the rocket exerts a force on the gases. Consider firing a rocket in space-there is nothing to “push against” there. 14. The two forces of the force pair act on different objects. 15. There are equal and opposite forces for all forces, and the net force is zero. 16. 9.8 N. Holding one end of the string before the pulley, the scale would measure only one mass.

Fr 2 N m2 Gm1m2 G = = , so, 17. F = m1m2 kg 2 r2 18. Because F  1 / r 2 , then F approaches zero only as r approaches infinity. 19. Smaller mass and size (radius). 20. Weight zero (gravity essentially zero). Same mass, 70 kg. 21. Only if the resultant of two or more gravitational forces of an object is zero. 22. Denser liquid, greater buoyant force 23. Enough volume for displacing sufficient water for floating. 24. Zero. Just another liter of water. 25. Same displacement, same force. 26. No. Large person displaces more water, but more buoyant force needed because of greater weight. 27. Balloon buoyant in air and would float away. 28. kg · m/s 29. In the absence of an unbalanced external force, an object or system has a constant velocity and hence a constant momentum. 30. (a) Gravity (the weight force) and the normal upward force of the surface on the blocks. (b) These forces cancel each other, and there is no net external force on the block.

Introduction to Physical Sciences 14e

31. Through the conservation of angular momentum. Tucking reduces the r of the mass distribution and the rotational speed increases.

ANSWERS TO VISUAL CONNECTION a. inertia, b. mass, c. constant velocity, d. net force, e. acceleration, f. m/s2, g. action, h. equal and opposite reaction, i. different objects

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Gravity is less (g/6), and in walking, the leg muscles are accustomed to applying a greater force for the Earth’s gravity. 2. Opposite reactions in both cases affects weight. (Downward force of arms, upward force on scale and more weight. 3. Set mg equal to Eq. 3.4. Moon has smaller M and R. 4. (a) To increase the amount of water displaced and increase the buoyancy. (b) Salt water is denser than regular water, and the average density of a person is less than that of salt water. 5. The adhesion between the water and the clothes is not sufficient to provide the necessary centripetal force for the water to rotate with the clothes, and hence the water becomes separated. 6. Shorter lever arm, less torque. 7. Shorter lever arm, more torque.

ANSWERS TO EXERCISES 1. (a) 3.0 N in direction of 8.0 N force. (b) 13.0 N in direction of forces. 2. 350 N to equal fmax. 3. F = m a = (3.0 kg)(5.0 m/s2) = 15 N 4. a = F/m = 2.1 N/(7.0  10–3 kg) = 300 m/s2 5. a = F/m = 950 N/1000 kg = 0.95 m/s2 6. a = F/m = 6.0 N/15 kg = 4.0 m/s2 7. w = mg = (6.0 kg)(9.8 m/s2) = 59 N 8. w = mg = (4.0 kg) (9.8 m/s2) = 39 N 9. (a) 120 1b (4.45 N/lb) = 534 N (b) Personal 10. (a) w = mg = (75 kg)(9.8 m/s2) = 74 N. (b) Same, zero acceleration. 11. F = Gm1m2/r2= (6.67  10–11 N-m2 /kg2)(3.0 kg)(3.0 kg)/(0.15 m)2 = 2.7  10–8 N

Introduction to Physical Sciences 14e

12. (a) F = Gm1m2/r 2= (6.67  10–11 N-m2/kg2)(103 kg)(103 kg)/(25 m)2 = 1.1  10–7 N (b) Much, much smaller; w = mg = (103 kg)(9.8 m/s2) = 9.8  103 N 13. (a) r2 = 2r1, and F 2 / F 1 = (r1/2r1)2= (1/2)2= ¼ (b) r2 = r1/2, and F 2/ F 1 = (2)2 = 4 14. (a) r2 = 2r 1/3, and F 2/ F 1 r1/r2)2 = (3/2)2 = 9/4 = 2.25 (b) r2 = 3r1 and F 2 F 1 = (1/3)2 = 1/9 15. (a) wM = wE /6 = 180 lb/6 = 30 lb (b) Personal. 16. g = F/m = 49 N/125 kg = 0.39 m/s2 17. Float. Density = m/V = 120 g/125 cm3 = 0.96 g/cm3, less than water. 18. Float. Density = 0.28 g/cm3 19. p = mv = (103 kg)(20 m/s) = 2.6  104 kg-m/s, east. 20. p = mv = (9.0  10 2 kg)(30 m/s) = 2.7  104 kg-m/s, north. Less, about 3/4 times less. Direction not a factor. 21. pg = –pb , or mgvg = –mbvb, and with m = w/g, vg = (-mb/mg)vb = (–wb/wg)vg = (–735 N/490 N)(0.50 m/s) = –0.75 m/s 22. mg = (0.75) mb , and vg = (–mb/mg)vb = (–1/0.75)(0.50 m/s) = –0.67 m/s 23. v2 = rlvl/r2 = (600  106 mi)(15,000 mi/h)/(100  106 mi) = 90,000 mi/h 24. (a) 36 km/h = 10 m/s. p = mv = (1.2 kg)(10 m/s) = 12 kg-m/s. (b) 76 mi/h = 33 m/s. p = mv = (1.2 kg)(33 m/s) = 40 kg-m/s.

Introduction to Physical Sciences 14e

Chapter 4

WORK AND ENERGY This chapter should be covered thoroughly in lecture and assignment. The relationship of work and energy is of the greatest importance in understanding many daily activities. Also, the development of the physical environment is closely associated with the control of energy. The law of conservation of energy is one of the most important general laws and has been a key to many of nature's secrets. Thus it is important for the student to know the meanings of work and energy and to be familiar with various forms of energy. Although this chapter deals primarily with general concepts and mechanical energy, it should be pointed out how easy it is to change other types of energy, such as chemical and electrical energy, to other forms and use them to do work. The textbook tries to get the student to think in terms of symbols, and this is a good chapter to stress this kind of thinking. For example, when referring to kinetic energy, think ½ mv2, and when thinking of gravitational potential energy, think mgh. In addition, the status of energy consumption and resources is covered, including alternative and renewable sources.

DEMONSTRATIONS A simple pendulum can be used to display the transformation of potential energy to kinetic energy, and vice versa. As the pendulum swings back and forth, ask the students at what stages the velocity, acceleration, potential energy, and kinetic energy have their minimum and maximum values. Demonstrate the examples of work as shown in the illustrations in the textbook. A radiometer can be used to show that light can do work. (General references to teaching aids are given in the Teaching Aids section.)

ANSWERS TO MATCHING QUESTIONS a. 14 b. 8, c. 13 d. 5

e. 9 f. 11 g. 1

j. 6

n. 10 o. 4

k. 7

l. 12

m. 2

h. 3 i. 15

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.a

2. d

3. d

4. c

5. b

6. d 7. c 8. a 9. b

10. a 11. d

12. c

13. b 14. c

Introduction to Physical Sciences 14e

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. parallel

2. scalar

7. transferring

3. joule

8. isolated

4. work

5. motion, position

9. power, watt

10. 0.75

6. square

11. energy

12. coal 13. exhausted 14.

ethanol (alcohol)

ANSWERS TO SHORT-ANSWER QUESTIONS 1. A force moving an object a distance. 2. No, there must be motion. No work is done in holding an object stationary, but work is done in lifting. 3. Friction. Dust reduces friction. 4. No work while stationary. Work was done on the weights in lifting. 5 Kinetic energy transferred to frictional heat. 6. B. vB = 2vA, mb = ma/3. KA = ½ mavA2

KB = ½ (ma/3)(2vA)2 = ½(4/3)(mavA2)

7. To take advantage of potential energy. 8. W2/W1 = (v2/v1)2 (2)2 = 4 9. Both may be correct, depending on the zero reference point chosen. 10. (a) The height depends on the initial kinetic energy, mgh = ½ mv2. (b) By the conservation of energy, it would have the same speed as it had initially. 11. Yes, relocate the arbitrary zero reference position. 12. Total energy includes all forms of energy. Mechanical energy is the sum of the kinetic and potential energies. 13. 50 J. Potential energy converted to kinetic energy. 14. Total energy: when energy neither enters or leaves a system and thus has a constant value. Mechanical energy: no energy loss. 15. (a) a and e. (b) c. (c) c. (d) a and e. (e) a and e. (f) c. (g) a and e. (h) c. (i) c. (j) a and e. 16. Same initial speed from same height. Both will have same speed on striking the ground. Conservation of energy. 17. (a) b and c. (b) a. (c) a. (d) b and c. (e) b and c. (f) a. (g) b, c; and e, d when going toward h = 0. (h) a; and e, d, when spring is compressing. (i) a; and e, d, when spring is compressing. (j) b, c; and e, d when going toward h = 0. 18. Yes, kinetic energy into potential energy into kinetic energy.

Introduction to Physical Sciences 14e

19. Yes, run the ¼-hp three times as long, or run larger motor one-third the time. 20. P = W/t, so person A, with the shorter time, expends more power. 21. (a) More work done in a given time. (b) Doing a given amount of work faster. 22. Energy, kilowatt-hour (kWh). 23. 100 W/60 W = 1.67 J (same time) 24. Oil. 25. Coal. 26. 100 J/s 27. Radiant, chemical, nuclear, sound, and heat. 28. Examples of alternative energy sources: electricity, natural gas, biodiesel, methane, ethanol, hydrogen, propane 29. Examples of renewable energy sources: hydropower, wind power, solar power, geothermal, tides 30. Solar and wind.

ANSWERS TO VISUAL CONNECTION a. work, b. energy, c. kinetic energy, d. power

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Yes, this is numerically possible. If there is no motion, the kinetic energy is zero. If the student selects his or her position as the zero reference point, then the potential energy is zero. 2. Same speed. Upward ball has same speed as other ball on return to initial height. 3. Body energy is used to increase height, but there is a limit. 4. Piecework involves power because the more energy expended per unit time, the greater the output and the more pay. An hourly rate implies a more constant power output, or at least compensation for same. 5. We can feel heat and vibrations, see energetic phenomena and light, and hear sound. However, it is doubtful that you could smell or taste energy. 6. Turn off lights when not needed, turn off appliances when not being used, keep thermostats properly set, and have good home insulation. (Limit children’s TV watching?)

Introduction to Physical Sciences 14e

ANSWERS TO EXERCISES 1. W = Fd = (250 N)(3.0 m) = 750 J 2. F = W/d = 400 J/2.0 m = 200 N 3. W = mgh = (5.0 kg)(9.8 m/s2)(0.45 m) = 22 J 4. W = mgh = (6.0 kg)(9.8 m/s2)(1.5 m) = 88 J 5. W = (60%) Fd = (0.60)(200 N)(6.0 m) = 7.2  102 J 6. W = (40%) Fd = (0.40)(200N)(6.0 m) = 4.8  102J 7. 14.7 J = W = mgh = (0.150 kg)(9.80 m/s2)(10.0 m) 8. 11.0 J = W = mgh = (0.150 kg)(9.80 m/s2)(2.50 m) 9. (a) Ek = ½ mv 2 = ½ (1000 kg)(25 m/s)2 =3.1  105 J (b) W = Ek = 3.1  105 J 10. Ek= ½ mv2, or v =

2E / m =

2(1.2x104J / 60kg = 20 m/s [36 km/h/(m/s)] = 72 km/h

[[ED. Space on each side of x and after 60.]] 11. Ek = ½ mv2 =½ (20 kg)(9.0 m/s)2 = 8.1  102 J 12. Ekb = ½mv2 = ½ (2.0  10-3 kg)(4.0  10 2 m/s)2 = 1.6  102 J. Eko = ½ mv 2= ½ (6.4  107 kg)(10 m/s)2= 3.2  109 J. Ocean liner has greater kinetic energy. 13. 32 J = K = W = Fd = (8.0 N)(4.0 m) 14. d = W/F = 42 J/8.0 N = 5.3 m, so 1.3 m farther. 15. Ep = mgh = (3.00 kg)(9.80 m/s2)(-10.0 m) = –294 J. Below ground zero point. 16. 294 J Ep = +294 J (same magnitude as in Exercise 15.) 17. ½ E p̂ (lost) = ½ mgh = ½ mg (12 m), and h = 6.0 m 18. (0.33E p̂ ) = (0.33) mgh = (0.33) mg (6.0 m), and h = 2.0 m 19. (a) E = mgh = (60 kg)(9.8 m/s2)(12 m) = 7.1  103 J. (b) Same. 20. mgh = ½ mv2, and v =

2gh (h from top) [[Ed. Square root sign over all]]

(a) v =

2 g (5.00 m) = 9.90 m/s

(b) v =

2 g (7.50 m) = 12.1 m/s

21. P = W/t = 7.2  10 2 J/10 s = 72 W 22. t = W/P = 7.2  102 J)/18W = 40 s 23. (a) W = Fh = (556 N)(4.0 m) = 2.2  103 J (b) P = W/t = 2.2  103 J/25 s = 89 W

Introduction to Physical Sciences 14e

24. P = W/t = mgh/t = 125 lb(1 kg/2.2 b)(9.8 m/s2)(4.0 m)/5.0 s = 4.5  102 W 25. E = Pt = (1.60 kW)(1/6 h) + (1.10 kW)(4/6 h) = 0.34 kWh (8¢/kWh) = 2.7¢ 26. (a) E = Pt = (1.25 kW)(4.0/60 h) = 0.083 kWh (b) 0.083 kWh (12¢/kWh) = 1¢

Introduction to Physical Sciences 14e

Chapter 5

TEMPERATURE AND HEAT Chapter 5 is an important chapter because temperature and heat are two of the most common physical concepts that students experience. In general, temperature measurements are given, and we say that heat is a form of energy. Hence it is important that a basic understanding and distinction of temperature and heat is obtained. In large part, the chapter is concerned with the measurement of macroscopic quantities of heat, such as specific heat and latent heat. The general trend is to express these heats in joules (J). However, calculations will be done primarily in kilocalories (kcal) because of the difficulty of adding numbers expressed in powers of 10, which is necessary when using joules. Calculations are much easier when done in kilocalories, and the results can be converted to joules if so desired. Because heat transfer has many applications in daily life, this is an important and interesting topic that should be covered in some detail. The chapter contains interesting Highlights: Freezing from the Top Down, and Hot Gases: Aerosol Cans and Popcorn. Finally, the basics of thermodynamics are discussed in Section 5.7.

DEMONSTRATIONS A thermometer may be calibrated in class by using boiling water for the steam point and ice water for the ice point. Uncalibrated thermometers are available, and students find it interesting and obtain a grasp for temperature scales when the interval between the ice and steam points is divided into degrees. (How should it be done?) Also, have a calibrated thermometer on hand so that you can check and see how accurate your calibration is. The bulb of one of two thermometers may be painted black and exposed to sunlight or a heat lamp to show the difference in radiation absorption.

ANSWERS TO MATCHING QUESTIONS a. 21 b. 19 n. 5

c.6

d. 22 e. 17

f. 9

g. 15

h. 25 i. 4

j. 11 k. 2

o. 13 p. 20

q. 18 r. 3

s. 24

t. 1

u. 12 v. 23

w. 14

l. 16 m. 7

x. 8 y. 10

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. c

2. a

3. a

4. c 5. b

6. c

7. a

8. b

9. c

10. c 11. c

12. b

Introduction to Physical Sciences 14e

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. greater 2. temperature 7. conduction

8. gas

3. 1000 4. J/ kg  C

9. Kelvin (absolute)

5. seven

10. inversely

6. pressure

11. direction

12. pump

ANSWERS TO SHORT ANSWER QUESTIONS 1. Fahrenheit 2. Celsius or Kelvin 3. Alcohol, low; mercury, high 4. Because of the thermal expansion of the bimetallic coil on which it sits. 5. Heat is energy in transit 6. Cold in, hot out 7. No 8. It is a measure of J/ kg C  for a particular substance and is characteristic of or specific for that substance. 9. Water in the filling has high specific heat 10. Half the water for the double temperature change 11. Condensation of moisture 12. Specific heat, c = J/ kg C  ; latent heat, J/kg. The latent heat process occurs at a particular temperature, hence there is no temperature change. 13. Thermal conductors: silver, copper, aluminum (metals). Thermal insulators: cloth, Styrofoam, wood. A difference in electron mobility and air space. 14. The tile floor which has a greater thermal conductivity. 15. Cold air underneath causes freezing 16. Loose fitting gives air insulation. Some have loose knitting for air spaces. 17. Temperature and pressure. 18. Solid: definite shape and volume. Liquid: definite volume, assumes shape of container. Gas: no definite shape or volume. (Volume may be restricted to a rigid container.) 19. (a) Sublimation. (b) Deposition. 20. Consists of molecules moving independently at high speeds in all directions. 21. A gas made up of point particles that interact only by collisions.

Introduction to Physical Sciences 14e

22. When the pressure is such that the space between molecules is small relative to the size of the molecules or the temperature drops to where attractions among the molecules are significant. 23. Molecular collisions with the walls of the gas container. 24. Frequent collisions exert a steady average force per unit area on the inside of the ball. 25. Heat is removed from the system (balloon), and negative work is done as the balloon collapses. 26. First law: Energy is conserved in thermodynamic processes. Second law: The direction of a process and whether or not a process will take place spontaneously. 27. First law (conservation of energy) and second law (entropy increases in every natural process). 28. It may be reduced by energy input, which requires a similar increase of entropy, so never destroyed. 29. It never decreases—the universe is the largest closed system of which we can think. 30. No, according to the third law of thermodynamics. 31. Zero.

ANSWERS TO VISUAL CONNECTION (a) deposition, (b) sublimation, (c) melting, (d) freezing, (e) condensation, (f) vaporization ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Hot air flows out by convection. 2. When steam condenses, latent heat is given up. 3. No, there is conduction (by coolant and through metal) and convection (by fan). 4. Initially the glass expands making the bore slightly larger. 5. Hole becomes larger. Would behave as circular cut out piece. 6. In a sense, the Earth absorbs energy (sunlight) and which does work (processes) in the environment. 7. No. Although there is more order and less entropy in the ice tray system, there is more disorder and a bigger increase in entropy somewhere else in the universe. 8. Answers may vary depending on approximations. Check and see how close.

Introduction to Physical Sciences 14e

ANSWERS TO EXERCISES 1. TF = (9/5)Tc + 32 = (9/5)17° + 32 = 63°F 2. (a) 245oF because of smaller degrees. (b) 375oF (200oC = 392oF) 3. TC = 5/9 (TF – 32) = 5/9(68° – 32) = 20°C 4. TC = 5/9 (TF – 32) = 5/9(103° – 32) = 39oC 5. (a) TC = 5/9(– 40° – 32) = – 40°C (b) TK = TC + 273 = 233 K 6. (a) TC = TK – 273 = 3 – 273 = –270 K (b) TF = (9/5)(–270) + 32 = –454°F 7. 100 kcal/h (4186 J/kcal) = 4.2 x 105 J/h = 420 kJ/h 8. 250 kJ (1 kcal/4.2 x 103 kJ) = 6.0 x 10-2 kcal 9. 0.45 Cal/ h·lb (8 h)(150 lb) = 540 Cal 10. 3500 Cal / (0.45 Cal/h) = 7.8  103 h (approx. 324 days, almost a year). 11. 3500 Cal / (325 Cal/h) = 11 h 12. 4.0 mi/h  11 h = 44 mi 13. H = mc  T = (0.50 kg)(1.0 kcal/ kg∙oC )(10OC) = 5.0 kcal 14. H = mc  T = (1.0 kg)(4186 kg∙oC )(100C°) = 4.2  105 J 15. (a) H = mc  T = (1.0 kg)(1.0 kcal/ kg∙oC ) (80 C°) = 80 kcal (b) (80 kcal) (0.00116 kWh/kcal)(12¢/kWh) = 1.1¢ 16. . H = mc  T. With H and m equal,  Ti/  Tal = cal/ci = (0.22 kcal/kg-oC)/(0.105 kcal/kg-oC) = 2.l.

So iron will have the higher temperature, 2.1 times higher. 17. HL = mc  T = (0.500 kg)(0.50 kcal/ kg∙oC )(10 C°) = 2.5 kcal kg)(80 kcal/kg) = 40.0 kcal,

H2 = mLf = (0.500

H3 = (0.500)(1.00 kcal/ kg∙oC )(20 C°) = 10.0 kcal

Total=

52.5 kcal 18. H1 = (0.200 kg)(0.50 kcal/ kg∙oC )(10 C°) = 1.0 kcal H2= (0.200 kg)(540 kcal/kg) = 108 kcal H3 = (0.200 kg)(1.0 kcal/ kg∙oC ) (100 C°) = 20 kcal

H4 = (0.200 kg)(80 kcal/kg) =

16 kcal Total = 145 kcal 19. With V2 = 4 Vij p2 = (V1/V2) p1 = (1/4) p1 20. p2 = (T2/T1) p2 = (1.2) p1 21. T1 = 20° + 273 = 293 K, T2 = 40° + 273 = 313 K

Introduction to Physical Sciences 14e

p2 = (T2/T1) p1 = (313 K/293 K)

p1 = (1.07) p1

22. T1 = 20° + 273 = 293 K T2 = (p2/p1) T1 = (1.5)293 K = 440 K; 440 K – 273 = 167°C 23. p2 = (V1/V2) p1 = (0.500 m3/0.150 m3)(200 Pa) = 667 Pa 24. T1 = 20° + 273 = 293 K T2 = (V2/V1) T1 = (0.600 m3)/0.500 m3)(293 K) = 352 K; 352 K – 273 = 79°C

Introduction to Physical Sciences 14e

Chapter 6

WAVES AND SOUND The concepts of waves, sound, and light are discussed in this chapter. Because most information about our environment comes to us by means of waves (see Section 1.3, “The Senses”), the general properties of waves are studied to prepare the student for the many physical concepts that involve waves. The general properties of waves are considered. Light is treated generally as waves, not electromagnetic waves since electric and magnetic fields have not been studied (Chapter 8). Major emphasis is given to sound waves because of their relationship to the environment. This includes the Doppler effect with its many practical applications. A qualitative discussion of resonance and standing waves without mathematics is presented.

DEMONSTRATIONS There are many demonstrations that illustrate waves, sound, and the Doppler effect. For the concept of transverse waves, a long length of rubber hose is very useful. The demonstration of longitudinal waves can be made with a toy Slinky. A number of demonstrations are available for sound and electromagnetic waves. Consult the Teaching Aids section (Appendix) for sources of these. Sound demonstrations are particularly easy and interesting. Illustrate the production of sound by several methods. Perform the ringingbell demonstration in a bell jar from which the air can be evacuated. The Doppler effect can be demonstrated using an ordinary hair dryer. Use an oscilloscope for graphic representation of different sounds, including speech. Standing waves can be demonstrated using a mechanical vibrator and a piece of string with a suspended weight.

ANSWERS TO MATCHING QUESTIONS a. 4

b. 7 c. 18

d. 15 e. 13 f. 9

g. 20

h. 12

i. 1

k. 14

l. 6 m. 16

n. 11 o. 3

j. 21

p. 8 q. 19

r. 5

s. 10 t. 17

u. 2

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b

2. a

3. a

4. a

5. d

6. b

7. c 8. b

9. d 10. a

11. b

12. a

13. d

Introduction to Physical Sciences 14e

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. energy

2. perpendicular

electromagnetic 9. intensity

3. wavelength

4. wavelength 5. light or 3.00  108 m/s

7. longitudinal 8. 20 kHz

10. 3 11. higher 12. approaching 13. natural or characteristic

ANSWERS TO SHORT-ANSWER QUESTIONS 1. A propagation of energy. 2. No, electromagnetic waves need no medium. Travels in vacuum. 3.(a) parallel to wave velocity direction. (b) perpendicular to wave velocity direction 4. (a) m (b) Hz (or 1/s)

(d) m

5. Two, the maximum wave displacement from the equilibrium position, and E  A2 6. (a) 2 cm (b) f = 1/0.6 s = 1.7 Hz 7. No, light travels at c in a vacuum. 8. Longer wavelength: red end. Higher frequency: blue end. 9. No, they are electromagnetic waves. 10. Light:  = 400 nm to 700 nm; sound:  = 10–1 m to 1 m. (Calculate, or from table.) 11. The lowest density part of a sound wave. 12. Loses energy, smaller amplitude. 13. 25,

1 25

14. (a) Frequency. (b) Intensity, or amplitude. (c) Harmonics, or overtones. 15. Sound from different parts of the band interfere. Interference distorts sound. 16. Energy is in joules (J). Intensity is J/s per m2 or W/m2. 17. No. The dB scale is not linear. (An increase of 3 dB doubles intensity.) 18. Light travels faster than sound. 19. (a) Shortened. (b) Lengthened. 20. (a) Shift to a higher frequency. (b) Shift to a lower frequency. 21. (680 m/s)/340 m/s = (Mach) 2 22. Detection involves the reflection of waves from an object. Speed and ranging involve the Doppler shift of the reflected waves. 23. Maximum energy transfer to the system when driven at a resonance frequency.

Introduction to Physical Sciences 14e

24. Node 25. The tension applied to a string. By varying the length (with a finger on the string) and/or tension (tuning).

ANSWERS TO VISUAL CONNECTION (a) transverse, (b) electromagnetic, (c) longitudinal, (d) sound, (e) inverse, f = 1/T, (f) wave speed ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Greater speed downwind. Slower speed against the wind. Strong wind only you may hear yourself. 2. No. There is no atmosphere on the Moon. Communication is by radio waves, which are electromagnetic waves that require no medium for propagation. 3. Through sunburns and suntans. UV also can cause cataracts and blindness (not a good detection method). 4. Faster than the speed of sound in water. 5. Because of standing waves (overtones) set up between the shower walls. 6. Open: L = 1/2, 3/2, and 5/2 wavelengths Closed: L = 1/4, 3/4, and 5/4 wavelengths

ANSWERS TO EXERCISES 1. T = 1/f = 1/5,0 Hz =0.20 s 2. T = 1/f = 1/(0,25  103 Hz) = 4.0 x10–3 s 3. (a) f = v /  = (2.0 m/s)/(1.5 m) = 1.3 Hz

(b) T = 1/f = 1/(1.3 Hz) = 0.77 s

4.  = c/f = (344 m/s)/(3000 Hz) = 0.115 m 5. (a)  = c/f = (3.00  108 m/s)/(6.50  10 5 Hz) = 4.62  102 m (b)  = (3.00  108 m/s)/(9.51  106 Hz) = 3.15 m 6.  = (3.00  l08 m/s)/(1018 Hz) = 3.00  10–10 m, or 0.300 nm 7. f = c/  = (3.00  108 m/s)/(420  10–9 m) = 7.14  1014 Hz 8. f = c/  = (3.00  108 m/s)/6.00  10–6 m = 0.500  1014 Hz From Figure 6.7 (in text), infrared region. 9. 9.48 x 1012 km . d = ct = (3.00 x 105 km/s)(3.16 x 107 s)

Introduction to Physical Sciences 14e

10. (a) d = 3000 mi = 4.8 x 106 m, t = d/c = (4.8 x 106 m)/3.0 x 108 m/s =1.6 x 10-2 s. (b) d = 3.50 x 105 m, t = d/c = (3.50 x 105 m)/3.0 x 108 m/s = 1.2 x 10-3 s 11.  = v/f = (344 m/s)/(50  103 Hz) = 6.9  10–3 m 12.  = v/f = (344 m/s)/(20 Hz) = 17 m  2 = 344 m/s/(2.0  104 Hz) = 0.017 m

13. vm =15(3344 m/s) = 5.2 x 103 m/s  = v m /f = (5.2 x 10 3 m/s)/20 x l0 3 Hz = 0.26 m 14. vm =  f =(0.333 m)(15.0 x 10 3 ) = 5.00 x 10 3 m/s. v m /v s = (5.00 x 10 3 m/s)/340 m/s = 150 times. 15. d = vt = (1/3 km/s)(4.5 s) = 1.5 km = (1/5 mi/s)(4.5 s) =0.90 mi 16. d = vt = (1/3 km/s)(9.0 s) = 3.0 km t = d/v s = (3.0 km)/(15 km/h) = 1/4 h = 12 min 17. 100 dB – 90 dB = 20 dB or 2 B, then10 2 or 100 times greater. 18. 10,000 = 104, and 4 B or 40 dB increase to 120 dB.

Introduction to Physical Sciences 14e

Chapter 7

OPTICS AND WAVE EFFECTS The six major properties of waves (reflection, refraction, dispersion, polarization, diffraction, and interference) are presented and discussed in this chapter. It is convenient to introduce the properties to the students by means of demonstrations. Reflection can be illustrated with a plane mirror. Also, note that everything in the classroom is seen by reflection except the light source. A pencil or a small ruler in a glass of water is a good demonstration of refraction, and a prism can be used to illustrate dispersion. Viewing a candle through the slit between two fingers is good for illustrating diffraction, as is viewing a candle flame through a feather. Interference of light waves is best illustrated by Young’s experiment. If an experiment with a spectroscope is not done in the laboratory, the instrument should be presented and demonstrated in class. Polarized light is easily demonstrated with crossed Polaroids. (Polarizing sunglasses may be used.) Considerable class time should be spent in constructing ray diagrams and locating real and virtual images for different types of spherical mirrors and lenses. To simplify matters, only raw diagrams are used to locate images. The mirror and lens equations have been omitted.

DEMONSTRATIONS There are many demonstrations for illustrating waves, sound, and the basic laws of optics. A few have been mentioned in the introduction above. The ripple tank is an excellent piece of apparatus for the production and projection of water waves. The apparatus can be used to study reflection, refraction, diffraction, and interference of waves. Distribute replica diffraction gratings to the students in class and let them see the spectrum from an incandescent light source. Demonstrate polarization with linear polarizers and double refraction with an Iceland spar crystal. If available, a laser is a spectacular demonstration in itself, and many optical properties can be demonstrated with commercial laser kits.

ANSWERS TO MATCHING QUESTIONS a. 15

b. 3 c. 18

d. 9 e. 7

f. 19

g. 20

h. 1 i. 11

j. 22 k. 5

l. 16

m. 10 n. 8

o. 17 p. 2

q. 6 r. 13

s. 21 t. 12

u. 14 v. 4

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d

2. d

3. b

4. a 5. b

6. c

7. b

8. c 9. c

10. a

11. a

12. a

Introduction to Physical Sciences 14e

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. geometrical or ray 6. converging 11. greater

2. specular, diffuse

7. cannot

8. thicker

3. vacuum 4. toward

9. concave or diverging

5. total internal

10. transverse

12. principle of superposition

ANSWERS TO SHORT-ANSWER QUESTIONS 1. (a) perpendicular back reflection to surface. (b) parallel to surface, no reflection. 2. Left, seen as right in mirror. (self-portrait) 3. Walk toward you at same speed. In-step but opposite feet (right-left reversal). 4. 12 in, but becomes smaller in appearance the farther from the mirror. 5. In mirror with right-left reversal 6. (a) no, perpendicular to surface. (b) no, parallel to surface. 7. No, above atmosphere 8. Refraction at water-glass and glass-air surfaces. 9. Because n = c/v, and c is greater than v. 10. Internal reflection and dispersion 11. Yes, sunlight refracted over horizon by atmosphere. No atmosphere, no refraction – shorter. 12. The distance between the vertex and the center of curvature is the radius of curvature R, and half this distance is the focal length, which defines the focal point at R/2. 13. Real images are formed on the object side of the mirror and can be displayed or focused on a screen. Virtual images are formed “behind” or “inside” the mirror, and cannot be formed on a screen. 14. (a) Concave: real, Do > f. (b) Convex: always virtual. 15. Reflected parallel to axis. 16. Through the focal point 17. To get a reflected beam 18. Concave 19. At the edges. 20. (a) Convex: real, Do > f; virtual, Do < f. (b) Concave: always virtual.

Introduction to Physical Sciences 14e

21. Because the lens or lens system gives an inverted image. Focusing involves adjusting the object distance so that a sharp image is formed on the screen. 22. The image is that of the Sun. The sunlight is concentrated on the spot. 23. To correct for different distances 24. Sunglasses and LCDs (for example). 25. No, sound waves are longitudinal and cannot be polarized. 26. Longitudinal waves cannot be polarized; transverse waves can be. 27. Yes, the sheets would be “crossed” and darken. 28. Use long wavelength of light compared to the size of an opening or object. Also use a diffraction grating. 29. The wavelengths of sound satisfy the diffraction condition. Light waves have very short wavelengths. 30. AM, longer wavelengths. 31. Interference.

ANSWERS TO VISUAL CONNECTION (a) object; (b) convex (converging); (c) real, (d) inverted, (e) reduced ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. A spherical convex mirror. 2. Reflections from different surfaces of glass. 3. The convex mirror produces a reduced image, and the smaller image may be interpreted as being more distant than it actually is. 4. (a) At infinity. (b) At the opposite focal point. 5. The fish would see the 360° above-water panorama in a circular cone defined by the critical angle. 6. Using polarizing sunglasses to see if you can darken the sunglasses with cross-polarization. 7. Darken (90°), (lighten 180°), darken (270°), and lighten (360°).

ANSWERS TO EXERCISES 1. θi = θr = 30° 2. θr = 90° – 30° = 60°

Introduction to Physical Sciences 14e

3. Bisecting triangles in the figure gives one-half height. Same for any distance. 4. 76 in./2 = 38 in., and 62 in./2 = 31 in. So,  = 7 in. 5. cm = c/ n = (3.00  108 m/s)/2.42 = 1.24  108 m/s 6. n = c/cm = (3.00  108 m/s)/(1.40  108 m/s) = 2.14 7. cm /c = 1/n = 1/1.52 = 0.658 (  100%) = 65.8% 8. cm /c = 0.413 = 1/n, and n = 1/0.413 = 2.42, diamond. (See Table 7.1.) 9. Sketch ray diagram. Real, inverted, same size. 10. Sketch ray diagrams. Reduced image becomes larger as objects move toward the mirror approaching the center of curvature, with M = 1 at that point. Continues to become larger as object approaches focal point, and image becomes virtual inside focal point. 11. Sketch ray diagram. 60 cm, virtual, upright, and smaller. 12. Sketch ray diagram. 5.5 cm, virtual, upright, and smaller. 13. Sketch ray diagram. Real, inverted, same size. 14. Sketch ray diagrams. Reduced image becomes larger as object moves toward lens approaching the 2f position, with M = 1 at that point. Then M > 1 as object approaches the focal point, and image becomes virtual inside the focal point. 15. Sketch a ray diagram. Di = 36 cm, real, inverted, and reduced. 16. Sketch ray diagrams. Di = 2f. 17. Parallel rays never meet or meet at infinity. No image. 18. Sketch a ray diagram. Di  60 cm, virtual, upright, and enlarged.

Introduction to Physical Sciences 14e

Chapter 8

ELECTRICITY AND MAGNETISM This is an important chapter because students should have a basic understanding of electric circuits because electricity plays such an important role in our lives. Also, knowledge of the fundamental concepts of electricity and magnetism is necessary in the study of atomic physics. Electric and magnetic laws and fields should be stressed so as to provide an understanding of some of the underlying concepts in the study of modern physics and chemistry. The discussion of electric fields has been expanded in the Thirteenth Edition, so as to make electromagnetism and waves better understood. This chapter is somewhat long, but the study of electric and magnetic phenomena is well worth the effort.

DEMONSTRATIONS Lecture demonstrations with an electroscope that can be projected on a large screen are very good and create interest. Demonstrate electromagnetic induction. (Commercial apparatuses are available.) Demonstrate the electric generator and electric motor. (May be constructed or commercially available.) Demonstrate the compass and dip needle. Pass around a cheap compass and bar magnets.

ANSWERS TO MATCHING QUESTIONS a. 6

b. 16 c. 8

m. 19

d. 24

n. 14 o. 7

e. 13 f. 20

p. 17

q. 9

g. 25

r. 15

h. 11 i. 23

s. 10

t. 1

u. 22

j. 2 k. 18

l. 5

v. 4 w. 12

x. 21

y. 3

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b

2. c

3. a

4. b

5. a

6. a 7. a 8. b

9. c

10. d 11. c

12. b 13. c 14. b

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. positively 7. I 2 R

2. amp-volt or (amp)2ohm

8. direct or dc 9. smallest

3. semiconductors

10. Curie

11. south

4. charge

5. open

6. ohm

12. secondary

ANSWERS TO SHORT-ANSWER QUESTIONS 1. Protons and neutrons. Protons and electrons.

Introduction to Physical Sciences 14e

2.Equal and opposite, Newton’s third law. 3. Because of attractive electrical forces arising from charging by induction and polarized molecules. 4. Static cling. 5. Electrical potential energy arises due to work done against an electric force. Voltage is electrical energy (or work) per unit charge, V = W/q. 6. A voltage source and a closed circuit. 7. P = V 2/R Small resistance, large joule heat, and vice versa. 8. Electric field. 9. Direct current has a voltage with constant polarity and flows in only one direction. Alternating current results from a changing polarity, and the current alternates in direction. 10. So that all will have the same voltage and independent paths so they can be operated independently. 11. (a) Opens a circuit by the melting of a metal strip when current is too large. (b) Opens a circuit by magnetic or thermal means when current is too large. (c) Dedicated grounding wire to prevent object being at high voltage. (d) Use of ground wire as a grounding wire. 12. Yes, less likely to strike. 13. Head-to-tail, series; all heads to tails, parallel. 14. Form small magnets that line up with the field. 15. Similar. Likes repel, unlikes attract. 16. (a) One that is easily magnetized. Iron, nickel, and cobalt. (b) The ferromagnetic material becomes an induced magnet. Above the Curie temperature, ferromagnetic materials loses their magnetism. 17. A current gives rise to a magnetic field in an iron core. 18. (a) The magnetic field of a bar magnet with the bar’s north magnetic pole near the Earth’s geographic south pole. (b) Declination is the horizontal angular distance between the Earth’s magnetic field lines or compass direction (magnetic north) and true north. Maps are based on true north, so it is necessary to know the declination to navigate properly. 19. Based on torque on a current-carrying wire. See text for description.

Introduction to Physical Sciences 14e

20. (a) Nothing. (b) Experiences a force at a right angle to motion, the direction given by the right hand rule. 21. Electromagnetic induction. Used to step-up and step-down voltage and current so as to reduce loss in electrical power transmission. 22. Much of the power would be lost through I 2 R losses. 23. Circuit not complete. 24. Voltage is the potential for high current and gets first warning.

ANSWERS TO VISUAL CONNECTION a. 1 A, b. 2 A, c. 2 A, d. 5 A. Combined resistance: 2.4 or 36/15 Ω

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Electrostatic charge, moisten. 2. (a) Midway for electron or proton, equal and opposite forces. (b) No. 3. (a) No (b) No 4. With one hand in the pocket, you are not likely to grasp a high voltage with both hands and provide a path across your chest. 5. (a) Air would oxidize filament. (b) Tungsten atoms has boiled off filament. 6. 1/Rp = 1/R1 + 1/R2 = (R2 + R1)/R1R2 and Rp = R1R2/(R1 + R2) 7. Two magnets would be obtained. Creating magnetic monopoles in this manner is not possible. Magnetism is the result of domain alignment. 8. At the location of the north magnetic pole, a dip needle would point vertically downward.

ANSWERS TO EXERCISES 1. n = q/e = 1.00 C/1.60  10–19 C/electron = 6.25  1018 electrons 2. q = ne = (106)(1.6  10–19 C) = 1.6  10–13 C 3. F = kq1 q2/ r 2 = (9.0  109 N m2 /C 2)(0.50 C)(2.0C )/(3.0 m)2 = 1.0  10–9 N, mutually repulsive. 4. F= k q1 q2/ r 2 = k (1.6  l0–19 C)2/(5.0  10–11 m)2 = 8.2  10–8 N F = Gmemp/ r 2 = (6.67  10–11N m2/kg2) (9.11  10–31 kg)(1.67  10–27 kg)/(5.0  10–11 m)2 = 4.1  10–47 N

Introduction to Physical Sciences 14e

5. I = q/t = ne/t = (4.8  1018)(1.6  10–19C)/0.25 s = 3.1 A 6. q = It = (1.50A)(6.0 s) = 9.0 C 7. U = W = 30 J 8. V = W/q = 30 J/0.25 C = 120 V 9. I = V/R= 120 V/50  = 2.4 A 10. V = IR = (0.50 A)(30  ) = 15  11. (a) P = IV = (0.25 A)(12 V)= 3.0 W (b) R = V/I = 12 V/0.25 A = 48  12. (a) I = P/V = 0.50 W/3.0 V = 0.17 A

(b) R = V/I = 3.0 V/0.17 A= 18 

13. E = Pt = (1.5 kW)(0.50 h/day)(30 day) = 22.5 kWh (  $0.08/kWh) = $1.80 14. E = (1.0 kW)(3.0 h/day)(30 day) = 90 kWh (  $0.10/kWh) = $9.00 15. P = V 2 / R = (12 V )2/24  = 6 W (6 J/s) 16. P = IV and V = IR or I = V/R, then P = (V/R)V = V 2/R 17. (a) R = V/I = 110 V/10 A = 11 

(b) P = IV = (l0A)(110 V)= 1100 W

18. (a) I = P/V = 100 W/120 V = 0.83 A

(b) R = V/I = 120 V/0.83 A = 144 

(b) P = V 2/RS = (12 V)2/60  = 2.4 W

20. RP = R1R2/(R1 + R2) = (25  )(35  )/50  = 15  21. (a) I = V/Rs = 50 V/45  = 1.1 A

(b) P = V 2/Rs = (50 V)2/45  = 56 W

22. 1/RP = 1/10  + 1/15  + 1/20  = (6 + 4 + 3)/60 = 13/60, and Rp = 60  /13 = 4.6  23. (a) 1/RP = 1/10  + 1/15  and Rp = 60  (b) R = 8  6.0  = 48  , and I = V/R = 120 V/48  = 2.5 A 24. (a) Rs = R1 + R2 + R3= 60  + 30  + 20  = 110  (b) Rp = R1R2/(R1 + R2) = (110  )(50  )/(110  + 50  ) = 34  I = V/R = 12 V/34  = 0.35 A 25. (a) V2 = ( N2 / N1 ) V1 = (300/100)(12 V ) = 36 V; (b) I2= ( N1 / N 2) V1 = (100/300)(2.0 A) = 0.67 A 26. (a) N1 > N2 step-down

(b) V2 = (200/500)(100 V ) = 40 V

I2 = (200/500)(0.25 A) = 0.63 A 27. N2 = (V2 / V1 ) N1 = (220 V/4400 V )(1000) = 50 turns

Introduction to Physical Sciences 14e

28. N2 = (40,000 V/240,000 V )(900) = 150 turns

Introduction to Physical Sciences 14e

Chapter 9

ATOMIC PHYSICS The subject matter of atomic physics includes some of the revolutionary concepts and developments of twentieth-century physics. First, for background, we review the models of the atom up to 1911. The dual nature of light is then introduced, accompanied by a Highlight on Einstein and his work. The next section discusses the Bohr theory of the hydrogen atom and includes a Highlight on fluorescence and phosphorescence. Quantum applications—microwave ovens, X-rays, lasers—are discussed. The Heisenberg uncertainty principle and the concept of the wave nature of matter are then explained, followed by a Highlight on electron microscopes. The chapter ends with a treatment of the quantum model of the atom. We have kept the discussions of these difficult topics as simple and straightforward as possible. Sections 9.1, 9.2, and 9.3 are basic material. Instructors should use their judgment as to what additional parts of the chapter to cover and emphasize.

DEMONSTRATIONS Many of the topics in this chapter are difficult to demonstrate. Fluorescence and phosphorescence may be demonstrated with an ultraviolet lamp and appropriate samples from supply houses. Students especially seem to like finding that fluorescent chalk has been used to write their homework assignment on the board. A common laser may be demonstrated, but use appropriate safety precautions. A food item may be heated in a microwave oven to demonstrate that the center remains cold if sufficient time is not allowed for conduction of heat to the center.

ANSWERS TO MATCHING QUESTIONS a. 11

b. 14

c. 5

k. 12

l. 16 m. 8

d. 16

e. 9 f. 6

n. 13 o. 3

g. 1

h. 15

i. 4

j. 2

p. 7 q. 10

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. c

2. a

3. a

4. a 5. c

6. d

7. b

8. d

9. d 10. a 11. b

12. d 13. d 14. d

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. electron 7. lower

2. Rutherford 8. water

3. Planck’s constant

9. unknown

10. uncertainty

4. photon 5. increases 11. wave

6. continuous

12. probability

Introduction to Physical Sciences 14e

ANSWERS TO SHORT-ANSWER QUESTIONS 1. Classical mechanics: macrocosm, motion of large objects. Quantum mechanics: microcosm, behavior of atoms and electrons on the atomic level. 2. Thomson found that electrons were deflected by electrical and magnetic fields in such a way that their electric charge had to be negative. 3. Thomson’s “plum pudding” model pictured the atom as consisting of electrons randomly positioned like raisins in an otherwise homogeneous mass of positively charged “pudding.” Thomson’s model was abandoned when Rutherford discovered that each atom had a tiny core, or nucleus, in which 99.9% of the mass and all the positive charge were concentrated and around which the electrons circulated. 4. Frequency increases with temperature. 5. The particle nature of light is shown by phenomena such as the photoelectric effect. 6. Something is quantized when it is restricted to certain discrete values rather than having a continuous range of values. 7. A proton is a positively charged subatomic particle found in the nuclei of atoms. A photon is a quantum, or “particle,” of electromagnetic radiation. 8. Frequency and wavelength are inversely proportional. 9. A photon of red light has less energy, lower frequency, and longer wavelength than a photon of violet light. 10. Only light above a certain frequency causes electrons to be ejected, and thus the photon must have a certain minimum energy to cause ejection; E = hf, the higher the frequency, the greater the photon energy. 11. Albert Einstein. The theory of relativity. 12. Same. 13. No, line spectrum – different transitions. 14. n 15. Bohr postulated that an orbiting electron does not radiate energy when in an allowed, discrete orbit but does so only when it makes a quantum jump, or transition, from one allowed orbit to another. 16. The ground state for an electron is the lowest energy state. Energy states above the ground state are called excited states.

Introduction to Physical Sciences 14e

17. Four visible lines—red, blue-green, and two violet—make up the line emission spectrum of hydrogen, as shown in Figures 9.9b and 9.13. 18. The four discrete lines in the emission spectrum of hydrogen correspond to electron transitions down to n = 2 from n = 6, 5, 4, and 3. 19. The four discrete lines in the absorption spectrum of hydrogen correspond to electron transitions up to n = 6, 5, 4, and 3 from n = 2. 20. n = 3 to n = 1 21. Three, 3 to 2, 2 to 1, and 3 to 1 22. The potato contains water, whereas the ceramic plate does not. 23. Light amplification by stimulated emission of radiation 24. Laser light is monochromatic, coherent, and has amplified intensity. 25. The intensity of laser light can cause eye damage. 26. X-rays are called braking rays because they are formed when high-speed electrons are stopped as they hit a metal plate. 27. It is impossible to know a particle’s exact position and velocity simultaneously. 28. Effects are negligible for everyday measurements 29. A moving particle has a wave associated with it called a matter wave, or de Broglie wave. The associated wavelength is significant only for atoms and subatomic particles. 30. Matter waves describe (fit) orbit. 31. Davisson and Germer showed that a beam of electrons undergoes diffraction, which is a wave phenomenon. 32. The electron microscope. 33. Schrödinger. 34. Quantum model, or electron cloud model.

ANSWERS TO VISUAL CONNECTION a. Dalton, billiard ball model; b. Thomson, plum pudding model; c. Rutherford, nuclear model; d. Bohr, planetary model; e. Schrodinger, electron cloud (or quantum) model

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Microwaves could damage the person operating the oven, because a person’s body contains water and could thus be “cooked” by stray microwaves.

Introduction to Physical Sciences 14e

2. A quantized cruise control might be digital and allow speeds to be set on 5 mph units, such as 50 or 55 mph, but not 51 mph. 3. As Highlight Figure 2 shows, individual atoms can be imaged and even moved around by use of a scanning tunneling microscope. 4. Mercury vapor or sodium vapor lights give off only certain wavelengths of light (those in their line emission spectra). Thus only certain colors are available to be absorbed or reflected from the variously colored cars. The reflected light by which we see the cars thus has a different overall composition (color) from reflected sunlight or reflected light from an incandescent bulb. 5. A car is much too large to allow  v (or  x) to be of significant size.

ANSWERS TO EXERCISES 1. E = h f = (6.63  10–34 J  s)(5.45  1014 1/s) =

3.61  10–19 J

2. E = h f = (6.63  10–34 J  s)(5.00  1014 1/s) = 3.32  10–19 J 3. (a) E = hf; f = (b) 

E 6.3  1019 J = = 1.00  1015 Hz 34  s h 6.63  10 J

c 3.00  108 m / s = = 3 . 0 0  10–7 m = 300  10–9 m = 300 nm f 1.00  1015 1 / s

4. (a) E = h f ; f =

(b)  =

E 1.50  1019 J = = 2.26  1014 Hz 34 h 6.63  10 J  s

c 3.00  108 m / s = = 1.33  10–6 m = 1330  10–9 m = 1330 nm 14 f 2.26  10 1 / s

5. rn = 0.053 nm  n 2 = 0.053 nm  9 = 0.48 nm 6. rn = 0.053 nm  n 2 = 0.053 nm  16 = 0.848 nm 7. En =

8. En=

13.60 13.60 eV = e V = –1.51 eV 2 n 32

13.60 13.60 eV = e V = – 0 . 8 5 eV 2 n 42

9. E photon = Eni – Enf = –0.85 eV – (–3.40 eV) = 2.55 eV 10. Ephoton = Eni –Enf = –13.60 eV – (–0.85 eV) = –12.75 eV (The minus sign indicates photon absorption.)

Introduction to Physical Sciences 14e

11.  =

h 6.63  1034 J  s = = 0.51  10–35 m mv (0.50kg)(26m/s)

12. Answers vary. (10 mi/h = 4.5 m/s, and convert mass, l lb = 0.45 kg)

=

h 6.63  1034 J  s = mv (mass, kg)(4.5m/s)

Introduction to Physical Sciences 14e

Chapter 10

NUCLEAR PHYSICS This chapter concludes the physics section of the textbook by discussing the atomic nucleus. Because of the great impact that nuclear energy has had—and will have—on society, this chapter is extremely important. The material in this chapter is so fundamental that we recommend it be covered in its entirety. Among the topics discussed are the names and symbols of elements, the structure and composition of nuclei, atomic mass, radioactive decay, half-life, nuclear reactions and reactors, fission, fusion, and the biological effects of radiation. The small amount of mathematics included is kept simple. Applications of radioactivity are stressed, and three Highlights emphasize the historical aspects of the discovery of radioactivity, the building of the first nuclear bombs, and the problem of nuclear waste disposal. Also in the Thirteenth Edition, a section on elementary particles has been added for those that wish to have an introduction to this topic. The student will need to refer frequently to the periodic table on the inside front cover of the textbook. We recommend that a wall chart of the periodic table be on display at all times in the lecture classroom, including during exams.

DEMONSTRATIONS Students generally find a demonstration of a Geiger counter or other radiation detector to be interesting and helpful. A comparison of the penetrating power of alpha particles, beta particles, and gamma rays is instructive (a solid source set #32852 can be obtained from CENCO or another science supply company). Another demonstration involves charging an electroscope and illustrating its quick discharge with a radioactive source. Of course, use appropriate safety precautions when dealing with radiation. An old X-ray of a human or animal obtained from a doctor or veterinarian may be passed around the class. A display Chart of the Nuclides is available at moderate cost from General Electric Company, 175 Curtner Ave., Mail Code 684, San Jose, CA 95125.

ANSWERS TO MATCHING QUESTIONS a. 26 b. 17 c. 25

d. 10

e. 5

f. 21

o. 22

r. 15

s. 7

t. 18 u. 16

p. 8 q. 11

g. 13

h. 9 i. 2

j. 20 k. 19

l. 3

v. 1

x. 4 y. 23

z. 12

w. 14

m. 24

n. 6

Introduction to Physical Sciences 14e

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b

2. d

3. c

4. d

5. c

6. d

7. b

8. c 9. c

10. c 11. c

12. a 13. c 14. b 15. a

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 2. 12C

1. phosphorus 8. one more

3. nucleons 4. isotopes

9. mass number

10. critical

5. 83

6. beta particle (electron) 7. four

11. deuteron (or deuterium)

12. 82

ANSWERS TO SHORT-ANSWER QUESTIONS 1. (a) C

(b) Cl

2. (a) Nitrogen.

3. Ru is taken for ruthenium. 4. The mass number is equal to the atomic number plus the neutron number, hence N = A – Z. 5. Nucleons. 6. Nucleus doesn’t fly apart. 7. That the nucleus occupies a very small core. 8. Over 99.9%. 9. Z is the atomic number, A is the mass number, and N is the neutron number. N = A  Z. 10. Protium (hydrogen), deuterium, and tritium. 11.

3 2 1 1H, 1H, 1H; protium (hydrogen), deuterium, and tritium

12. Carbon-12, which is assigned a mass of exactly 12 u. 13. The strong nuclear force (or just nuclear force). It drops to zero at distances greater than about 10–14 m. 14. Transformation into different nucleus. 15. Proton number increases by one. 16. Less by mass of 2 protons and 2 neutrons. 17. (a) Gamma.

(b) Beta.

(d) Gamma.

(e) Alpha.

18. One-eighth. 19. Activity becomes too weak and unable to count. 20. a + A  B + b 21. Ionized air attaches to smoke particles affecting monitoring circuit. 22. Using neutron beam to form isotopes. Isotopes can be identified through gamma decay. 23. Neutron. 24. The minimum amount of fissionable material necessary to sustain a chain reaction is called the critical mass. A subcritical mass would be less than that, and no chain reaction would be

Introduction to Physical Sciences 14e

possible. A supercritical mass would be more than that, and an expanding chain reaction would be possible. 25. 0.7%, which must be enriched to about 3% for use in a U.S. nuclear reactor, and to 90% or more for use in nuclear weapons. 26. Fusion of fuel rods. Melting through reactor floor. 27. Control rods adjust the number of neutrons available to cause fission. Moderators slow down fast neutrons from fission so that they can more effectively cause other fissions. 28. For a nuclear explosion to occur, the fissionable material must be of much greater purity than the fuel in nuclear reactors. 29. No. Nucleus is “split” into two nuclei. 30. Plutonium-239 is made from uranium-238. 31. Hydrogen 32. D is a deuterium nucleus, or deuteron; T is a tritium nucleus, or triton. 33. A plasma is a very hot gas of electrons and protons or other nuclei. In magnetic confinement, electric fields are used to form the plasma, and electric and magnetic fields are used to confine its charged particles. The second process is inertial confinement using lasers. 34. Fusion has the advantages of low cost and abundance of fuel, fewer nuclear waste disposal problems, and the impossibility of a runaway accident. Its disadvantages are that it has not yet proved practical, and that fusion plants will be more costly to build and operate than fission plants. 35. Einstein. E is energy, m is mass, and c is the speed of light. 36. High temperature needed to start fusion reaction. 37. Exoergic, mass defect. 38. Sketches. 39. Radiation that can ionize atoms or molecules to form ions. Such radiation can kill living cells. 40. In order of needing more shielding: alpha particles, beta particles, and gamma rays, X-rays, and neutrons. 41. Exchange particles are responsible for force interactions. 42. Three quarks (two with +2/3 and one with -1/3 electronic charge).

Introduction to Physical Sciences 14e

ANSWERS TO VISUAL CONNECTION (a) 73Li4, (b) 126C6, (c) 168O8, (d) 2311Na12

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Charged particles would be deflected in different directions. (See Chapter 8.5.) 2. More intensity, greater concentration of carbon-14. Samples would be older. 3. Answers will vary according to the individual, but major considerations would be that coalburning plants release more radioactivity than do nuclear plants during normal operation, plus the emissions of coal-burning plants would cause much grime to collect on and in your house and could well adversely affect your respiratory system. On the other hand, an accident in a nuclear plant might have more devastating effects than one in a coal-burning plant. 4. Answers might include not smoking, fewer high-altitude airplane flights, living closer to sea level, watching TV from a greater distance, fewer X-ray diagnoses, and radon checks of your crawl space or basement. 5. Provide fundamental knowledge about the makeup of matter.

ANSWERS TO EXERCISES 1. Fill in the nine gaps in this table. Symbol

Protons

Neutrons

Electrons

Mass number

2. Fill in the nine gaps in this table. Symbol

Protons

Neutrons

Electrons

Mass number

Introduction to Physical Sciences 14e

16 17 18 8O8, 8O9, 8O10

48 19 190 4. 63Li3, 80 35Br45, 22Ti26, 8F11, 78Pt112

228

5. (a)  , gamma

(b) 88 Ra, alpha

(c)

0 e, beta 1

6 (a) 8 C, beta 6

(b) 4 He, alpha 2

226 222 7. (a) 98 Ra → 86 Rn + 4 He 2 (b) 60 Co → 60 Ni + 0 e 27

1

225

221

8. (a) 89 Ac → 87 Fr + 4 He 2 221

221

(b) 87 Fr → 88 Ra + 0 e 1 218 4 9. (a) 222 86Rn → 84Po + 2He 214 4 (b) 218 84Po → 82Pb + 2He 218 0 218 84Po → 85At + −1e

10. Determine isotope from atomic and proton numbers. 249

11. (a) 98 Cf (Z > 83) 76

(b) 33 As (odd-odd) 18

12. (a) 9 F (fewer n than p) 226

(b) 88 Ra (Z > 83) 20

36 h = 6 half-lives 6.0 h/half-life

Introduction to Physical Sciences 14e

14. 160  80 cpm  40 cpm  20 cpm  10 cpm  5 cpm Five arrows are shown, so five half-lives are needed. 15.

24.3 d  3 half-lives 8.1 d/half-life 1  1/2  1/4  1/8

16.

75 h = 5 half-lives 15 h/half-life

480 cpm  240 cpm  120 cpm  60 cpm  30 cpm  15 cpm will be the activity. 17. 1  1/2  1/4  1/8  1/16 Four arrows mean four half-lives have elapsed. (4 half-lives)(12.3 y/half-life) = 49 y 18. 2000  1000  500  250 Three arrows mean three half-lives.

21 min  7.0 min/ half-life 3 half-lives 19. 22 days 20. 8 days 21 . about 38 days 22. 30 days 23. (a) 1 H 1 (b) 1 n 0

2 (c) 1 H (d) 93 Sr 38 24. (a) 24 Mg 12 (b) 28 Al 13 (c) 1 H 1 (d) 254 No 102 25. 3 1 n 0

Introduction to Physical Sciences 14e

26. 96 Tc 43 27. 3  4.00260 u = 12.00780 u on left – 12.00000 u on right = 0.00780 u mass loss (0.00780 u)(931 MeV/u) = 7.26 MeV or energy produced 28. 2.0140 u + 2.01401 u = 4.0280 u showing on the left. 3.0160 u + 1.0078 u = 4.0239 u showing on the right. The difference is 0.0041 u more on the left, so mass is lost and (0.0041 u)(931 MeV/u) = 3.8 MeV of energy produced.

Introduction to Physical Sciences 14e

Chapter 11

THE CHEMICAL ELEMENTS Chapter 11 begins the four-chapter chemistry section by discussing the chemical classification of matter, the discovery and occurrences of the elements, the periodic table, and the basics of compound nomenclature. A chapter Highlight discusses Berzelius and How New Elements Are Named. Conceptual questions and answers are presented on the uniqueness of compounds as well as on why a “periodic table of compounds” is impractical. Much of the material discussed in this chapter is necessary for dealing successfully with the next three chapters. The student already should have learned the names and symbols of the 46 elements in Table 10.2 of the textbook and will now need to learn the names and formulas of the eight polyatomic ions listed in Table 11.6 and the eleven common compounds shown in Table 11.3. Probably the most efficient and effective way of doing this is to make and use flashcards.

DEMONSTRATIONS When discussing the differences between compounds and mixtures, it is useful to pass around two Erlenmeyer flasks, one containing the compound zinc sulfide and the other containing a heterogeneous mixture of powdered sulfur and mossy zinc. Solubility can easily be demonstrated with a beaker, salt, and water. Try changing the temperature of the water to see how much more salt can be dissolved. Elemental abundance in Earth’s crust can be demonstrated by looking at the chemical formulas for common igneous, sedimentary, and metamorphic rocks in Chapter 22, and specifically in Table 22.8.

ANSWERS TO MATCHING QUESTIONS a. 20 b. 17 c. 9 d. 11 e. 19 f. 5 g. 6 h. 15 i. 16 j. 3 k. 23 l. 25 m. 24 n. 22 o. 4 p. 8 q. 2 r. 10 s. 7 t. 1 u. 18 v. 21 w. 13 x. 14 y. 12

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b 2. c 3. a 4. b 5. d 6. c 7. c 8. c 9. b

10. d 11. a 12. c

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS

Introduction to Physical Sciences 14e

1. Organic chemistry 2. solvent 3. more 4. technetium (Tc) 5. molecules 6. aluminum 7. graphite 8. Mendeleev 9. four 10. increases 11. acetate 12. sodium hydroxide

ANSWERS TO SHORT-ANSWER QUESTIONS 1. Chemistry deals with the composition and structure of matter and the reactions by which substances are changed into other substances. 2. Physical, analytical, organic, inorganic, and biochemistry. 3. Illustrations b and d show different atoms and/or molecules and thus represent mixtures. Illustration e shows identical molecules composed of the same two elements and so represents a compound. Illustrations a and c show identical atoms and identical diatomic molecules of atoms of the same element, respectively, and thus represent elements. Illustration c shows diatomic molecules of an element, whereas e shows diatomic molecules of a compound. 4. A homogeneous mixture, featuring a type of solid solution, results when copper and zinc are mixed to form brass. 5. Mixtures can be separated into, and prepared from, pure substances by physical processes. 6. An element is a pure substance in which all the atoms have the same number of protons; elements cannot be further broken down by chemical processes. A compound is a substance composed of two or more elements chemically combined in a definite, fixed proportion by mass. 7. Compounds can be prepared from, and decomposed into, elements by chemical processes only. 8. An appropriate set of examples would be: bronze, salt water, air. 9. Carbonated beverages are bottled and capped tightly to maintain storage under high pressure. Once a carbonated beverage is opened, the pressure inside the bottle is reduced to normal atmospheric pressure over time, and the CO2 starts escaping from the liquid, eventually resulting in the flat taste. 10. Any six of these: gold, silver, lead, copper, tin, iron, carbon, sulfur, antimony, arsenic, bismuth, or mercury. Most of these are metals. The first gases isolated were hydrogen, oxygen, nitrogen, and chlorine. 11. The first synthetic element, technetium (Tc) was produced by nuclear bombardment of molybdenum with deuterons. 12. At present, about 118 elements are known, 88 of which occur naturally on Earth.

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13. (a) Oxygen and silicon. (b) Iron and nickel. (c) Oxygen and carbon. (d) Nitrogen and oxygen. (e) Hydrogen and helium. 14. The three noted allotropes of carbon are diamond, graphite, and fullerenes. In diamond, each carbon atom is in the middle of a geometric structure called a regular tetrahedron, where the four bonds of each carbon atom point toward the four corners of the tetrahedron, resulting in a threedimensional network that helps make diamond the hardest substance known. Graphite consists of carbon atoms that are bonded in a network of flat hexagons, where each carbon atom is bonded to three other carbon atoms that lie in the same plane, forming sheets of flat, interlocking hexagons. Fullerenes consist of soccerball-like arrangements of interlocking hexagons and pentagons formed by carbon atoms. 15. Dmitri Mendeleev developed the periodic table in 1869. 16. Atomic number is now used. The periodic trends in the table are based on atomic number, which correlates more exactly with electron configuration than does atomic mass. 17. Periodic characteristics of elements include: 1) atomic number, 2) electron configuration, 3) valence electrons, 4) atomic sizes, and 5) ionization energy. One can also include metallic character in this list. 18. (a) Periods. (b) Groups. 19. The valence electrons are the ones that form chemical bonds. 20. (a) Metals usually have 1 to 3 valence electrons; nonmetals usually have 4 to 8. (b) Metals are good conductors of heat and electricity; nonmetals are not. (c) Metals are solid at room temperature (except for Hg); many nonmetals are gaseous or liquid. 21. Metallic character (a) decreases from left to right across a period, and (b) increases down a group. 22. Semi-metals are those elements that display both metallic and nonmetallic characteristics (i.e. they only conduct electricity under certain conditions). Examples are boron, silicon, and germanium; they are typically used in the semiconductor industry for electrical circuits.23. Mercury and bromine are liquid. Hydrogen, oxygen, nitrogen, fluorine, and chlorine, (plus six noble gases) are gases. 24. Atomic size (a) decreases from left to right across a period, and (b) increases down a group. 25. Methane, nitrous oxide, nitric oxide, and sulfuric acid, respectively.

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26. (a) An atom is the smallest particle of an element; a molecule is an electrically neutral particle composed of two or more atoms chemically bound together. (b) An atom is electrically neutral; an ion has an electrical charge. (c) A molecule is an electrically neutral particle composed of two or more atoms; a polyatomic ion is an electrically charged particle composed of two or more atoms. 27. The Greek prefixes designate the number of atoms of the element that occur in the molecule. 28. Because they have the same number of valence electrons. 29. Noble gases have interesting properties in that they are not very reactive. That is, it is monotonic, and it does not bond easily with other elements. Their valence electron shells are completely full, and thus are stable. They have very low melting and boiling points. Noble gases are used in lighting in two ways: they are used in incandescent bulbs to insulate the hot tungsten filament from other reactive gases, and they can be used in “neon” signs giving an emission spectrum. 30. Fluorine and chlorine are gases, bromine is liquid, and iodine is solid. Fluorine is the most reactive element. 31. Sodium iodide is added to prevent thyroid problems due to iodine deficiency. 32. K2CO3, Na2CO3, NaOH, and Mg(OH)2, respectively. 33. Common uses for alkaline earth metals and their compounds include: safety flares, metal alloys, medicines, acid neutralizers, road materials, chalk, and colors and dyes. 34. Hydrogen is considered a nonmetal, not a metal. However, hydrogen appears in Group IA because it usually reacts similar to other alkali metals.

ANSWERS TO VISUAL CONNECTION a. pure element, b. mixture of element/cpd, c. mixture of element/cpd, d. pure compound, e. pure element, f. mixture of two elements

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Element 118 would belong to Group 8A and Group 7 – it would resemble the noble gases. Element 116 would belong to Group 6A and Group 7 – it would resemble elements like oxygen, sulfur, and polonium. 2. The letter J.

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3. Although homogenized milk looks uniform to the unaided eye, a microscope shows fat globules dispersed throughout the water. Therefore, it is not a true solution, because it is not mixed on the atomic or molecular level. 4. Graphite makes a good lubricant because of its slipperiness, so the key should turn easily. 5. Being “snobs” refers to the noble gases’ lack of chemical reactivity, or “interaction,” with other elements. 6. Fluorine is the most reactive element of all and thus is so dangerous that it should be handled only by experts. 6. For this hypothetical case, Ca-20 would have 20 protons and 10 electrons. No, neutral hydrogen needs 1 proton and 1 electron, each with the same charge. 7. ClO2

ANSWERS TO EXERCISES 1. (a) Homogeneous mixture. (b) Compound. (c) Element. (d) Heterogeneous mixture. 2. (a) element, (b) compound, (c) homogeneous mixture, (d) heterogeneous mixture 3. About 65 g/100 g H2O. 4. About 2½ times (interpolated from graph). 5. Unsaturated; about 240 g of sugar are soluble per 100 g H2O at 40°C. 6. Saturated; interpolation from graph shows that only about 130 g of NaNO3 are soluble in 100 g H2O at 60°C. 7. (a) S (b) Na (c) Al 8. (a) Pt, (b) Po, (c) Ra. 9. (a) Nitrogen. (b) Potassium. (c) Zinc. 10. (a) Antimony, (b) Arsenic, (c) Mercury 11. (a) 3, 2A (b) 4, 2B (c) 5, 4A 12. (a) Group 7A and Period 6, (b) Group 2A and Period 5, (c) Group 3A and Period 5. 13. (a) Representative, nonmetal, gas. (b) Transition, metal, solid. (c) Inner transition, metal, solid. 14. (a) Inner transition/metal/solid, (b) transition/metal/solid, (c) representative/nonmetal/gas. 15. (a) 14, 4, 3 (b) 33, 5, 4 16. (a) 4, 2, 2 (b) 16, 6, 3

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17. (a) 6.94 u, 3, 3, 3 (b) 197.0 u, 79, 79, 79 18. (a) 39.9 u, 18, 18, 18 (b) 87.6 u, 38, 38, 38 19. (a) 2 (b) 2,8,3 20. (a) 2,1 (b) 2,8,5 21. (a) Ca, Mn, Se (b) Po, Se, O 22. (a) Na, P, Cl (b) Br, Cl, F 23. (a) Sr, Sn, Xe (b) Ar, Ne, He 24. (a) Cs, K, Na (b) Ca, As, Br 25. (a) Kr, Br, Ca (b) Li, Rb, Cs 26. (a) Ne, C, Be (b) Si, Ge, Pb 27. (a) Sulfuric acid. (b) Nitric acid. (c) Hydrochloric acid. 28. (a) Phosphoric acid. (b) Acetic acid. (c) Carbonic acid. 29. (a) Calcium bromide. (b) Dinitrogen pentasulfide. (c) Zinc sulfate. (d) Potassium hydroxide. (e) Silver nitrate. (f) Iodine heptafluoride. (g) Ammonium phosphate. (h) Sodium phosphide. 30. (a) Aluminum carbonate. (b) Ammonium sulfate. (c) Lithium sulfide. (d) Sulfur trioxide. (e) Barium nitride. (f) Barium nitrate. (g) Silicon tetrafluoride. (h) Disulfur dichloride. 31. Li2S, Li3N, LiHCO3 32. Ba(NO3)2, BaCl2, BaSO4, and Ba3(PO4)2 and Al(NO3)3, AlCl3, Al2(SO4)3, and AlPO4

Introduction to Physical Sciences 14e

Chapter 12

CHEMICAL BONDING This chapter outlines the early history of chemistry and discusses the law of conservation of mass, the law of definite proportions, and Dalton’s atomic theory. The chapter’s first Highlight focuses on Lavoisier, the father of chemistry. The chapter’s second Highlight discusses sunglasses and photochromic lenses. The basic ideas of ionic, covalent and polar covalent, and hydrogen bonding are discussed. Review electric charges and electric forces if necessary (Chapter 8.1). After the student is able to determine ionic charges from chemical formulas, the Stock system of nomenclature is introduced. Conceptual questions and answers are giving to help understand covalent, ionic, and hydrogen bonding. The material in this chapter should be covered in its entirety. For success in Chapters 12 through 14, the student must understand the trends of ionic charge and number of covalent bonds formed for atoms of the representative elements, as given in Figure 12.7 and Table 12.4, and must develop the ability to write the formulas of ionic and covalent compounds using the ideas presented in Sections 12.4 and 12.5.

DEMONSTRATIONS Large models of NaCl, CaCO3, and solid CO2 are helpful when discussing the structure of ionic and molecular solids. These are available from sources such as Klinger. Have the students build various models of the atom (see Chapter 9.1) and discuss the properties of each. Remind students the scale of the atom compared with the size of the nucleus (Chapter 10.2). Have the students calculate the size of a hydrogen atom using a peanut as the nucleus.

ANSWERS TO MATCHING QUESTIONS a. 7

b. 16

c. 2 d. 9

e. 13

f. 3 g. 15 h. 8

i. 11 j. 19

k. l l. 5 m. 12 n. 17 o. 4 p. 14 q. 6 r. 18 s. 10

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b 2. a 3. d 4. c 5. d 6. a 7. c 8. c 9. d 10. a 11. c 12. d 13. a 14. b

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ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. mass 2. CO2 3. definite proportions 4. chemistry 5. CuO 6. anion 7. two 8. M2X 9. ionic 10. double 11. decreases 12. nitrogen 13. higher, 14. lower, 15. greater

ANSWERS TO SHORT-ANSWER QUESTIONS 1. The law of conservation of mass states that no detectable change in total mass occurs during a chemical reaction. For example, if a sealed vessel containing copper and oxygen has a mass of 245.00 g and is heated so that copper oxide is formed, the vessel and its contents will still have a mass of 245.00 g after reaction. 2. Lavoisier discovered the law of conservation of mass, explained the role of oxygen in combustion, established the principles for naming chemicals, wrote the first modern chemistry textbook, and introduced quantitative methods into chemistry (four of the five are requested). 3. Joseph Priestley discovered oxygen, which he called “dephogisticated air.” 4. The isotope is 12C and its assigned mass 12 u. 5. Dry ice is solid carbon dioxide, which is very cold (–78°C) and sublimes (passes directly from the solid phase to the gaseous). 6. The reactant used completely is the limiting reactant (B in this example). A is the excess reactant. 7. Dalton hypothesized (1) that each element is composed of small indivisible particles called atoms, which are identical for that element but different from atoms of other elements, (2) that chemical combination is the bonding of a definite number of atoms of each of the combining elements to make one molecule of the formed compound, and (3) that no atoms are gained, lost, or changed in identity during a chemical reaction. 8. The percentage abundance of isotopes of a given element is the same in all natural samples on Earth. This means that each element has an unvarying average mass (the atomic mass) that can be assigned to its atoms. The law of definite proportions holds whether atoms of a given element all have the same mass or an unvarying average mass. 9. In forming compounds, atoms tend to gain, lose, or share valence electrons to achieve the electron configurations of the noble gases. That is, they tend to get eight electrons (an octet) in the outer shell (except for H, which tends to get two electrons in the outer shell, like the

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configuration of He). Hydrogen is an exception because it must use the first shell, which can hold only two electrons. 10. The valence electrons, those in the outermost shell of the atom. 11. Atoms with three or fewer electrons tend to lose them; atoms with four or more valence electrons tend to gain more electrons. 12. Sodium bonds with chlorine via an ionic bond. Sodium will lose its valence electron to form a sodium ion of charge +1 that has a stable shell (octet) of electrons. This extra electron is used by a chlorine atom to form a stable chloride ion with charge -1. Electrical forces attract these ions together to form a strong, ionic bond. The summary reaction is this: Na+ + Cl- → NaCl 13. Two K+ ions and one S2– ion. 14. The nucleus and all the inner shell electrons. The number of dots is the same as the atom’s number of valence electrons. 15. A Lewis symbol shows the valence electrons of a single atom or ion. Lewis structures show the valence electrons in the molecules and ion combinations that make up compounds. 16. Metals form positive ions, whereas nonmetals form negative ions. Cations are positive ions; anions are negative. 17. Isoelectronic means having the same electron configuration (in this case, two electrons in the first shell and eight in the second). 18. The total electric charge must be zero, and all atoms must have noble gas configurations. 19. If it conducts electricity when molten, it is ionic. 20. In the solid phase, ions are held in place and thus cannot move, thereby unable to conduct an electric current. 21. The Stock system is preferred whenever a metal that can form more than one type of ion is part of the compound. 22. Covalent compounds are formed by the sharing of pairs of electrons. 23. A single bond is one shared pair of electrons between two atoms, whereas double and triple bonds result from the sharing of two and three pairs, respectively. 24. Electronegativity increases across (left to right) in a period and decreases down a group. 25. Fluorine has the highest electronegativity, and oxygen has the second highest. 26. A polar covalent bond.

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27. No; because there is only one polar bond, the polarity could not be offset by an opposing polar bond. 28. The carbon and oxygen atoms within the polyatomic ion are covalently bonded, while the whole CO32- aggregation behaves like an ion (due to the overall -2 charge) when forming ionic compounds. 29. No. Carbon tetrachloride is a compound of two nonmetals; therefore it uses covalent bonds. Materials with covalent bonds are poor conductors of electricity in all forms. 30. Covalent. With a boiling point of –10°C, it is a gas at room temperature, and ionic compounds are not gaseous at room temperature 31. The lone pair of electrons on the nitrogen central atom results in a trigonal pyramidal geometry with an overall molecular dipole according to VSEPR. 32. Like dissolves like, which means that polar liquids tend to dissolve polar and ionic compounds but not covalent compounds, whereas covalent liquids tend to dissolve other covalent compounds but not polar and ionic substances. 33. When water freezes, the alignment of the hydrogen bonds between the water molecules in ice produces a more open structure, thereby increasing its volume and making its density less than liquid water.

ANSWERS TO VISUAL CONNECTION a. trigonal planar

b. angular

c. trigonal pyramidal

d. tetrahedral

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. The lowest common denominator of 12 and 8 is 24. To get 24 buns would require the purchase of two packages; to get 24 wieners, you would purchase three packages. This problem is similar to finding the correct ratio of positive and negative ions needed to give net neutrality and thus to find the correct formula for an ionic compound. 2. The law of conservation of mass states that matter cannot be created or destroyed. Although pollutants may be burned and changed into other substances, no total mass is lost, so waste products are still present. Dissolving in water may only disperse the original molecules and not change them at all.

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3. To completely use the 200 bodies would require 800 wheels. You have 900 wheels on hand, so 200 cars can be produced, and 100 wheels will be left over (in excess). This problem is similar to having an excess of one reactant in a chemical reaction. 4. Vinegar is polar and oil is nonpolar, so a heterogeneous mixture results, with the less dense oil on top. Shaking causing a dispersion (intermixing) so that some of each component pours out, if you act fast! 5. Photochromic lenses darken and lighten as the ambient light changes. These lenses have a tiny amount of silver chloride crystals to the glass. External light causes an electron transfer between a chloride ion and the silver ion forming neutral silver and chlorine atoms. The flakes of silver metal in the glass are opaque to light and thus darken the glass. The ionic reaction is Ag+ + Cl→ Ag + Cl. Amazingly this process is reversible, but needs other ionic reactions with copper ions to work. See the Highlight: Automatic Sunglasses in Chapter 12.4 for a complete set of ionic reactions. 6. There are a few exceptions to the trend that electronegativity decreases down a group. Some are gallium, indium, and thallium in Group 3A. Others are germanium, tin, and lead in Group 4A. These elements have smaller atomic size than is expected based on the trends in their individual Groups and Periods (see Fig. 11.23). The smaller size allows the nucleus to draw bonding electrons to it. The nuclei in larger-sized atoms are more shielded and thus have lower electronegativity values. Notice that some of these exceptions are seen in the ionization energies (Fig 11.24).

ANSWERS TO EXERCISES 1. 68 tons + 96 tons – 36 tons = 128 tons 2. 0.942 g + 56.413 g – 57.136 g = 0.119 g 3. (a) 12.0 u + (2× 16.0 u) = 44.0 u (b) 12.0 u + (4× 1.00 u) = 16.0 u (c) (3 × 23.0 u) + 31.0 u +(4 × 16.0 u) = 164.0 u 4. (a) 180.2 u, (b) 342.3 u, (c) 749.0 u 5. 100.0% – 25.5% = 74.5% 6. 27.8% nitrogen (N) 7. (a) (23.0 u/58.5 u) × 100% = 39.3% Na ; (35.5 u/58.5 u) × 100% = 60.7% Cl (b) 42.1% C , 6.4% H , 51.5% O (FM = 342.0 u)

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8. (a) 40.1% Ca, 12.0% C, 48.0% O (FM = 100.1 u) (b) 41.7% Mg, 54.9% 0, 3.4% H (FM = 58.3 u) 9. (a) (6.1 g/14.1 g) × 100% = 43% Mg (b) 14.1 g – 6.1 g = 8.0 g S , law of conservation of mass. 10. (a) (7.75 g/67.68 g) × 100% = 11.5% P (b) 67.68 g – 7.75 g = 59.93 g Br, law of conservation of mass. 11. Still 14.1 g of MgS, because 6.1 g of magnesium will react with only 8.0 g of sulfur, leaving 2.0 g of sulfur in excess; law of definite proportions. 12. Still 67.68 g of PBr3, because 59.93 g of bromine will react with only 7.75 g of phosphorus, leaving 2.25 g of phosphorus in excess; law of definite proportions. 13. (a) 2– (b) 1+ (c) 1– (d) 3– (e) 2+ (f) 0 (g) 0 (h) 3+ 14. (a) -1, (b) +1, (c) -2, (d) +2, (e) +3, (f) -3, (g) 0, (h) +4 ..

215. Na+ Na+ :O: ..

16. Ba2+ :Br: . . :Br: ..

17. (a) CsI (b) BaF2 (c) Al(NO3)3 (d) Li2S (e) BeO (f) (NH4)2SO4 18. (a) RbI, (b) Cs3P, (c) Li2SO4, (d) Ag2CO3, (e) Zn3(PO4)2, (f) Al(NO3)3 19. (a) Iron(II) hydroxide. (b) Copper(II) chloride. (c) Gold(III) iodide. 20. (a) Zinc nitrate. (b) Chromium(II) sulfide. (c) Copper(I) oxide. 21. (a) 2 (b) 0 (c) 1 (d) 3 (e) 4 22. (a) 1 (b) 0 (c) 1 (d) 4 (e) 3 ....

23. H:N:N:H .... HH

24.

.. H:C:O .. H

.. .. H—N—N—H | | H H .. H—C—O | H

25. (a) HCl (b) NC13 (c) SC12 (d) CC14 26. (a) CH4 (b) CBr4 (c) CS2 (d) No compound formed. 27. (a) Covalent, two nonmetals. (b) Ionic, Group 1A metal and nonmetal. (c) Ionic, metal and polyatomic ion. (d) Covalent, two nonmetals. (e) Covalent, all nonmetals. (f) Ionic, two polyatomic ions. 28. (a) Covalent, two nonmetals. (b) Ionic, Group 1A metal and nonmetal.

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(c) Covalent, two nonmetals. (d) Ionic, Group 2A metal and nonmetal. (e) Ionic, metal and polyatomic ion. (f) Covalent, three nonmetals. 29. (a) S and Cl are in the same period, but Cl is farther right than S, so Cl is the more electronegative, and the arrows would point to the two Cl atoms. (b) C and O are in the same period, but O is farther right and so is the more electronegative. Thus the arrows would point to the two oxygen atoms. 30. (a) Cl is higher than Br and in the same group, and so Cl is the more electronegative atom, and the dipole arrow will point toward the Cl. (b) N is higher than I, but I is farther right. Figure 12.12 shows that N has EN = 3.0, whereas I has EN = 2.5. So N is the more electronegative atom, and the dipole arrow in each bond will point toward the N. 31. (a) linear, polar; (b) trigonal pyramidal, polar; (c) angular, polar; (d) trigonal planar, nonpolar 32. (a) linear, polar; (b) tetrahedral, nonpolar; (c) trigonal planar, nonpolar; (d) trigonal pyramidal, polar

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Chapter 13

CHEMICAL REACTIONS This chapter discusses the differences in chemical and physical changes, how and why equations are balanced, the role of energy in chemical reactions, factors that influence the rate of reactions and the basic types of chemical reactions (combination, decomposition, hydrocarbon combustion, single-replacement, and double-replacement). The students are made aware of chemical reactions that occur in their everyday lives. Acids, bases, and pH are discussed, and the concepts of oxidation and reduction are explained. The chapter’s final topic is a discussion of Avogadro’s number, the mole concept and examples of mole to gram conversions. Chapter Highlights discuss the chemistry of tooth enamel and the chemistry of airbags. Conceptual questions and answers are given to help understand the rates of reaction and acids and bases. We recommend that all sections of this chapter be covered. The completing and balancing of equations that illustrate a few basic reaction types is the most important skill developed by the student in this chapter. Information and skills attained in Chapters 11 and 12 will be used extensively.

DEMONSTRATIONS Any demonstrations of chemical reactions will be valuable aids to student understanding and appreciation. Examples include the combination reaction when magnesium burns, a combustion reaction involving burning ethanol, an acid-carbonate reaction such as CaCO3 and HCl, a single replacement reaction such as placing a zinc strip in aqueous CuSO4, or (carefully) demonstrating the reaction of a small piece of Zn with dilute hydrochloric acid. Another good demonstration would be the MnO2-catalyzed decomposition of KClO3, but take appropriate safety precautions, including a safety shield. An excellent demonstration of an exothermic reaction is to place KMnO4 crystals to a height of about one inch in a 16 × 150 mm test tube, place the test tube in an Erlenmeyer flask, and add about seven drops of glycerol. After a few seconds, smoke will start to rise, and very shortly a flame will shoot up in the tube. It is advisable to use a safety shield.

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The difference between a physical change and a chemical change can be demonstrated by mixing powdered sulfur and iron filings. That this is only a physical change is shown by using a magnet to separate the iron filings from the sulfur. Each substance keeps its own characteristic properties, and no chemical combination has occurred. If the mixture is placed in a test tube and heated, the iron and sulfur combine to give FeS. Now a magnet will not separate the iron from the sulfur.

ANSWERS TO MATCHING QUESTIONS a. 26 b. 24 c. 18 d. 15 e. 12 f. 7 g. 2 h. 5 i. 11 j. 23 k. 3 l.21 m. 8 n. 16 o. 10 p. 25 q. 17 r. 6

s. 20 t. 4 u. 19

v. 22 w. 14 x. 1 y. 9 z. 13

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d 2. d 3. b 4. b 5. d 6. a 7. c 8. a 9. c 10. b 11. a 12. d

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. physical 2. the same 3. AB 4. A + B 5. lower 6. catalyst 7. higher 8. neutral 9. acid-carbonate 10. redox 11. salt 12. 2

ANSWERS TO SHORT-ANSWER QUESTIONS 1. The reactants are carbon dioxide and water. The products are oxygen and glucose, the catalyst is chlorophyll, and the energy source is the Sun. 2. That it is blue-black, crystalline, solid, and that it sublimes are all physical properties. That it reacts with aluminum and many other metals is a chemical property. 3. (a) Physical. (b) Physical. (c) Chemical. (d) Chemical. 4. Reactants disappear, new substances (products) appear, and energy is released or absorbed. 5. In a chemical reaction the reactants are diminished (used up partially or completely) to form the products. These products have different chemical properties than the reactants. For the reaction to take place energy in the form of heat, light, electricity, or sound is either given to the system (absorbed) or released from the system. 6. In a chemical reaction the coefficients represent the relative number of substances in the reaction. These numbers can be changed to correctly balance the equation. The subscripts,

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however, represent the absolute amount of atoms or molecules in each substance. These subscript numbers are fixed and cannot be changed. 7. Do not start with oxygen, because it is present in two places on the product side. 8. (a) Fractional coefficients are generally not used. (b) The smallest set of whole numbers has not been used; 4, 2, 4 should be changed to 2, 1, 2. (c) Hydrogen should be written H2, not H, on the product side. (The correct set of coefficients would then become 2, 2, 2, 1.) (d) Helium is nonreactive. 9. Aqueous, or water, solution; solid; liquid; gas; reacts to give; reversible reaction, or equilibrium; catalyst. 10. (a) 2 H2O → 2 H2 + O2 -- this is a decomposition reaction (b) 2 Ca + O2 → 2 CaO -- this is a combination reaction (c) 2 NaN3 (s) → 2 Na (s) + 3 N2 (s) -- this is a decomposition reaction 11. The reaction vessel feels warm for an exothermic reaction because energy is being liberated (released) during the reaction. An endothermic reaction, however, requires energy input to the system therefore, the reaction vessel will feel cold. 12. Energy. 13. (a) Reactants. (b) Products. (c) Activation energy. (d) Heat of reaction. (e) Exothermic. 14. Carbon dioxide and water are the products, and oxygen is the other reactant. If O2 is in short supply, carbon monoxide and carbon are also products. 15. For reaction to occur, the collision must occur at the proper places in the molecules (proper orientation) and must have at least enough energy to break bonds (the activation energy). 16. Temperature, concentration, state of subdivision, and catalyst. 17. (a) An increase in temperature increases the frequency of collisions and their average energy. (b) An increase in reactant concentration means that collisions will be more frequent. 18. Powdered zinc and powdered sulfur are in finer states of subdivision than the lumps, and so more reactant surface is available for collisions. 19. A catalyst speeds up the rate of the reaction by providing a new pathway with lower activation energy.

Introduction to Physical Sciences 14e

20. A catalytic converter changes the poisonous carbon monoxide (CO) and nitric oxide (NO) gases coming from a car’s engine into less harmful carbon dioxide (CO2) and harmless nitrogen (N2). 21. 7; 1000 times. 22. Blue, red. 23. pH meter; the ammonia solution, at the left. 24. Acids: conduct electricity, turn blue litmus red, taste sour, react with bases to neutralize their properties, react with active metals to liberate H2. Bases: conduct electricity, change litmus from red to blue, react with acids to neutralize their properties. 25. Arrhenius acids are substances that give hydrogen ions, H+ (or hydronium ions, H3O+), in water. Arrhenius bases give hydroxide ions, OH–, in water. 26. (a) Hydrochloric acid, HCl. (b) Magnesium hydroxide, Mg(OH)2. (c) Sodium hydroxide, NaOH. (d) Sodium bicarbonate (sodium hydrogen carbonate), NaHCO3. (e) 5% acetic acid, HC2H3O2. 27. Water and a salt. The most common mistake is writing the wrong formula for the salt. 28. Hydrates. 29. Water, carbon dioxide, and a salt. 30. AB+CD→AD+CB 31. Particles of solid form in the mixture of the two clear solutions as the positive ions of one solution combine with the negative ions of the other to form an insoluble salt. 32. Oxidation is a gain of oxygen or a loss of electrons. Reduction is a loss of oxygen or a gain of electrons. 33. The relative activity of a metal indicates its tendency to lose electrons to ions of another metal or to hydrogen ions. 34. Hydrogen gas and a salt. This is a single-replacement reaction. 35. Gold and silver are not very active. Sodium and magnesium are active metals, high in the activity series, and so form compounds readily. 36. There is Avogadro’s number of formula units in a mole. 37. The molarity is the number of moles of solute per liter of solution. 38. Molarity is defined to be the ratio of the number of moles of solute per volume of solution. To calculate the number of moles of salt in a solution, multiply the molarity and the volume.

Introduction to Physical Sciences 14e

ANSWERS TO VISUAL CONNECTION a. combination b. AB → A + B c. single replacement d. AB + CD → AD + CB e. hydrocarbon combustion ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. a is a combination reaction(x); b is a combustion reaction(y); c is a double replacement reaction(v); d is a decomposition reaction(u); e is a single replacement reaction(w). 2. Acidic 3. Charcoal burned indoors would liberate a dangerous amount of carbon monoxide gas. 4. The citric acid dissolves and ionizes, and the hydrogen ions react with the sodium hydrogen carbonate to liberate CO2 gas—the bubbles. 5. In the body, enzymes catalyze the reactions; thus activation energy requirements are lower. 6. Sticks have more surface area than the same mass of logs, so they react more rapidly when heated in the presence of oxygen. 7. Baking soda is an excellent absorber of gases that can cause unpleasant odors. 8. The baking of bread. 9. The water comes from the combustion of the gasoline. Hydrocarbon burning gives CO2 and H2O. 10. Strong acids ionize readily; that is, they give hydrogen ions (H+) in solution. This means that the valence electrons of the compound are easily removed, and therefore, the ionization energy is very small. The valence electrons “live” higher in the potential energy well (see Fig 9.11) and it takes less energy to remove them from the well. Weak acids have their valence electron deeper in the well and require more energy to become ionized. Thus they do not become ionized as readily in solutions.

ANSWERS TO EXERCISES 1. (a) 1, 1, 1, 2 (b) 4, 3, 2 (c) 2, 3, 1, 3 (d) 1, 2, 1, 1 (e) 2, 2, 1 (f) 2, 15, 12, 6 2. (a) 2, 1, 2 (b) 4, 3, 2, 6 (c) 2, 13, 8, 10 (d) 2, 2, 4, 1 (e) 8, 3, 4, 9 (f) 1, 2, 1, 1 3. (b) Combination. (e) Decomposition. 4. (a) Combination. (d) Decomposition. 5. (a) N2 + 3 H2 → 2 NH3 (b) 2 KCl → 2 K + Cl2

Introduction to Physical Sciences 14e

6. (a) Ca + Cl2 → CaCl2 (b) Na2CO3 → Na2O + CO2 7. C5H12 + 8 O2 → 6 H2O + 5 CO2 8. 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O 9. (a) HNO3 + KOH → H2O + KNO3 (b) 2 HC2H3O2 + K2CO3 → H2O + CO2+ 2 KC2H3O2 (c) H3PO4 + 3 NaOH → 3 H2O + Na3PO4 (d) H2SO4+ CaCO3 → H2O + CO2 + CaSO4 10. (a) 2 HCl + Ba(OH)2 → 2 H2O + BaCl2 (b) 6 HCl + Al2(CO3)3→ 3 H2O + 3 CO2+ 2 A1C13 (c) H3PO4 + 3 LiHCO3→ 3 H2O + 3 CO2 + Li3PO4 (d) 2 Al(OH)3 + 3 H2SO4 → 6 H2O + A12(SO4)3 11. (a) AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) (b) BaC12(aq) + K2CO3(aq) → BaCO3(s) + 2 KCl(aq) 12. (a) K2SO4(aq) + Pb(NO3)2(aq) → PbSO4(s) + 2 KNO3(aq) (b) 2 K3PO4(aq) + 3 CaBr2(aq) → 6 KBr (aq) + Ca3(PO4)2 13. (a) Will not. (b) Will. (c) Will. 14. (a) Will (Zn is above Fe). (b) Will not (Pb is below Fe). (c) Will not (Ag is below H). 15. (a) Ni + Pt(NO3)2(aq) → Pt + Ni(N03)2(aq) (b) Zn + H2SO4(aq) → H2 + ZnSO4(aq) 16. (a) Mg + 2 HCl → H2 + MgC12 (b) Al + 3 AgNO3(aq) → 3 Ag + Al(NO3)3(aq) 17. 12.04 × 1023 molecules. 18. 24.08 × 1023 molecules. 19. 6.02 × 1023 atoms, 23.0 g; 3.00 mol, 69.0 g; 3.00 mol, 12.04 × 1023 atoms. 20. 18.06 × 1023 molecules, 132 g; 1.00 mol, 6.02 × 1023 molecules; 0.50 mol, 22.0 g. 21. 185.25 g 22. 0.094 mol 23. 2.7 x 1021 molecules 24. 9.40 x 1024 molecules 25. 721 grams 26. 1.15 x 103 g

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Chapter 14 ORGANIC CHEMISTRY This chapter deals with the division of chemistry that probably most affects our daily lives. Organic compounds are all around us and, in fact, basically are us. The chapter begins with a discussion of the bonding rules for organic molecules. Aromatic and aliphatic hydrocarbons are then treated, followed by a section on derivatives of hydrocarbons and synthetic polymers. The last section is an introduction to biochemistry including brief discussions of carbohydrates, fats, and proteins. The two chapter Highlights discuss some relatively new and interesting products in organic chemistry. The conceptual questions and answers illustrate one use of polymers and the structure of some biochemical molecules. For success in handling this chapter, the student must learn at once the basic bonding rules in Table 14.1 (these also were covered in Chapter 12) and the names and molecular formulas of the first eight alkanes (Table 14.2). In order for the student to understand simple isomers, the instructor must continually emphasize the tetrahedral geometry of the four single bonds to carbon and that no difference exists in drawing a bond to carbon “up,” “down,” or “sideways” on the two-dimensional blackboard. Some instructors may not feel comfortable teaching organic chemistry, but every effort has been made to make the presentation of the chapter material clear, consistent, and coherent. If time is short, the instructor can stop at any point in the chapter but should not skip a section to go to one further on, because each section builds on the preceding section.

DEMONSTRATIONS The use of ball-and-stick models to show bonding is recommended, especially for the aromatic hydrocarbon compounds. Also, the naming of organic compounds and the structure of isomers are most easily discussed with the aid of such models.

ANSWERS TO MATCHING QUESTIONS a. 7 b. 17 c. 13 d. 23 e. 16 f. 15 g. 20 h. 3 i. 24 n. 14 o. 25 p. 5 q. 19 r. 8 s. 1 t. 6

j. 26 k. 9 l. 18 m. 11

u. 22 v. 4 w. 10 x. 12 y. 21 z. 2

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ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b 2. c 3. d 4. c 5. b 6. b 7. a 8. d 9. a 10. c 11. c 12. b

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. Carbon 2. three 3. Covalent 4. aliphatic 5. hydrogen 6. cycloalkane 7. alkyne 8. alcohols 9. amide 10. condensation 11. Teflon 12. carbohydrate

ANSWERS TO SHORT-ANSWER QUESTIONS 1. Organic chemistry studies the compounds that contain carbon. Biochemistry studies the chemical compounds and reactions that occur in living cells. 2. Carbon (4 bonds), hydrogen (1 bond), oxygen (2 bonds), sulfur (2 bonds), nitrogen (3 bonds), a halogen (1 bond). 3. A compound with the formula CH5 would necessarily have 5 bonds to a carbon atom, which is not possible. 4. Benzene is an aromatic hydrocarbon; it is a clear, colorless liquid, and as the name indicates, it has a very strong odor. The molecular formula is C6H6. 5. Kekulé ,

preferred

. 6. An aromatic hydrocarbon contains one or more benzene rings. 7. Alkanes, CnH2n + 2, CH3CH3, etc. Cycloalkanes, CnH2n, etc. Alkenes, CnH2n, CH2 = CH2, etc. Alkynes, CnH2n - 2, CH2 ≡CH2, etc. 8. Methane, CH4 (the principal component of natural gas) Ethane, C2H6 Propane Butane Pentane Hexane Heptane Octane

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9. The bonds point toward the corners of a regular tetrahedron. They form angles of 109.5° to one another. 10. Constitutional isomers are compounds that have the same molecular formula but different structural formulas. 11. For organic compounds, the basic rules of nomenclature are: 1) the longest chain of carbon is found, and the compound is named as a derivative of the alkane, 2) the positions and names of the substituents replacing hydrogen on the longest chain are added, and 3) the carbon atoms are numbered starting at the end of the chain closest to the substituents. 12. The full and condensed structural formulas for methane, methyl group, ethane and ethyl groups are: H

H–C–H

H–C–

Methane, CH4

methyl group, CH3 –

H H

H–C–C–H

H–C–C–

H H

Ethane, CH3CH3

ethyl group, CH3CH2 –

13. Benzene and Cyclohexane are structurally different. Benzene is an alkene with 3 doublebonded carbons in a ring (condensed formula, C6H6), while cyclohexane is a saturated hydrocarbon ring with only single-bonded carbon atoms (condensed formula, C6H12). 14. Cyclopentane, C5 H10,

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15. Ethylene, H2C = CH2. Acetylene, HC ≡CH. 16. Saturated hydrocarbons are those in which the hydrogen content is at a maximum. Unsaturated hydrocarbons have double or triple bonds, which allow the addition of more hydrogen atoms. 17. An alkene can add a molecule of H2 to the double bond. An alkyne can add two molecules of H2 to the triple bond. Alkanes cannot undergo addition reactions because to do so would make five bonds to a carbon atom, which can form only four. 18. CH3OH 19. Chlorofluorocarbon. Used in air conditioners, refrigerators, and heat pumps. Upon reaching the stratosphere, CFCs provide chlorine atoms that can destroy ozone molecules that protect life by absorbing UV radiation. 20. Alcohols have the general formula R-OH, contain one or more hydroxyl groups (-OH), and have names ending in -ol. 21. Amines have the general formula R-NH2, contain one or more amino groups (-NH2), and have unpleasant odors. 22. The general formula for a carboxylic acid is RCOOH. It contains a carboxyl group (-COOH). 23. Esters are popular because they have pleasant odors. They are found in oranges and other ripe fruits, and in the fragrances of flowers. The general formula for an ester is RCOOR'. 24. Amides are nitrogen-containing compounds and have the general formula R–CONH2. They are used as solvents and plasticizers. 25. Addition polymers are formed when alkene monomers add to one another, so a carboncarbon double bond is necessary. 26. Condensation polymers are constructed from molecules that have two or more reactive groups. Each molecule attaches to two others by ester or amide linkages.

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27. Dacron is a polyester, and nylon is a polyamide. Common addition polymers include polyethylene, Teflon, Styrofoam, and polyvinyl chloride. 28. Humans get energy from carbohydrates by breaking bonds in digestion to form simple sugars which can be used in cells to produce energy. The three main types of carbohydrates are sugars, starches, and cellulose, although humans cannot digest cellulose. Plants make carbohydrates by using energy from the Sun in photosynthesis. 29. (a) Fats and oils are triesters of glycerol.(b) Proteins are polyamides. 30. Glucose and fructose combine to form sucrose. Glucose. 31. Herbivores have bacteria in their digestive tracts that have the enzymes necessary to break the glucose linkages in cellulose. 32. Amino acids. The simplest is glycine,

33. Saturated fats are chemically different from unsaturated fats. Saturated fats contain all singlebond carbon atoms that are saturated with hydrogen, whereas unsaturated fats have double bonds between some of the carbon atoms (i.e. are not saturated with single carbon-hydrogen bonds). Physically, saturated fats are in solid form at room temperature while unsaturated fats are liquid at room temperature (also called oils). The saturated fats are dangerous in large quantities because they can lead to cholesterol buildup in arteries. 34. Soap is composed of the sodium salts of the fatty acids produced when fats are treated with sodium hydroxide.

ANSWERS TO VISUAL CONNECTION a. alkene

b. ester

c. amine

d. amide

e. carboxylic acid

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. The element silicon, Si, is in the same group and is a nonmetal like carbon. However, its bonding capabilities do not come close to those of carbon.

Introduction to Physical Sciences 14e

2. “Cat cracking” is short for catalytic cracking, the production of smaller hydrocarbons from larger ones by use of a catalyst. Many oil refineries are located in Louisiana. 3. Aspirin is an aromatic hydrocarbon (specifically, a carboxylic acid). It contains a carboxyl group, – COOH, and a hydroxyl group, – OH. Acetaminophen is also an aromatic hydrocarbon (an amide). It contains an amide group, – NCO, and a hydroxyl group, – OH. 4. Yes, a “2” inside the triangle indicates high-density polyethylene, which can be recycled (Section 14.5). 5. Peel the onion under the surface of water in a bowl. 6. Water and soap are used to clean a spot of grease off of clothing. Water is a polar solvent and soap is a long-chain sodium salt of fatty acids; soap is polar at one end of the long molecule and nonpolar at the other end. Water dissolves the polar end of the soap molecule, and the grease (a nonpolar solvent) dissolves the other end of the soap molecule. The grease is suspended in the water-soap mixture (emulsified) and is swept away by rinsing. 7. Hydrogen bonds exist between some of the H atoms and O or N atoms in other parts of the same protein molecule that have twisted around into close proximity. The result is that proteins twist themselves into long curls and then fold onto themselves. ANSWERS TO EXERCISES 1. (a) Is valid. (b) Is incorrect because Cl should have one bond, not two. 2. Formula a is incorrect because the C has only three bonds when it should have four. 3.

5. (a) Alkene. (b) Cycloalkane. (c) Alkyne. (d) Alkane. (e) Aromatic.

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6. (a) Aromatic. (b) Alkane. (c) Cycloalkane. (d) Alkene. (e) Alkyne. 7. (a) Alkane. (b) Alkyne. (c) Cycloalkane. (d) Alkene. (e) Aromatic. 8. (a) Alkene. (b) Cycloalkane. (c) Alkane. (d) Alkyne. (e) Aromatic. 9. (a) Same compound. (b) Neither. (c) Isomers. 10. (a) Neither. (b) Same compound. (c) Isomers. 11. (a)

(b)

12 (a)

(b)

13. 1-pentene CH2CHCH2CH3 and 2-pentene CH3CHCHCH2CH3. (3-Pentene would be the same as 2-pentene, and 4-pentene the same as 1-pentene.)

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14. Three isomers: 1-hexene, CH2CHCH2CH2CH2CH3 2-hexene, CH3CH = CHCH2CH2CH3 3-hexene, CH3CH2CH = CHCH2CH3 15. (a) Amine. (b) Amide. (c) Ester. (d) Carboxylic acid. (e) Alkyl halide. (f) Alcohol. 16. (a) Carboxylic acid. (b)Ester. (c) Alcohol. (d) Amine. (e) Alkyl halide. (f) Amide. 17. (a) CH3CH2-OH and CH3-O-CH3 (b) CH3CH2CH2Cl and CH3CH(Cl)CH3 18. (a) CH3CH2NH2 and CH3NHCH3 (b) CH3CH2CH2OH CH3CH(OH)CH3 CH3OCH2CH3 19.

20.

21.

22.

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23.

24.

CH2 O

25.

26.

O C11H23

C O

CH2

C11H23

C O

C11H23

CH2

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Chapter 15 PLACE AND TIME This chapter is very important for the prospective elementary teacher, as well as other students. The authors have found that the topics of this chapter, which are so common environmentally, are of strong interest to students. In this chapter, we examine how events are located, or their positions designated, in space and time. Cartesian coordinates are introduced, and the concept of a grid expanded to locations on the spherical Earth in terms of latitude and longitude. There is a Highlight on global positioning, daylight saving time, and the calendar. How time is measured is of major importance, and the measurement units of days, hours, standard time zones, daylight savings time, and so on, are discussed. The year is related to the seasons, and the chapter ends with the historical development of our calendar.

DEMONSTRATIONS With colored markers, draw the equator, small circles parallel to the equator, the prime meridian, and one or two other meridians on a 16-in.-diameter sphere. Show the sphere axis tilted 23° down from the vertical. Place a light source representing the Sun at the center of the lecture desk; then move the slate sphere around the light, illustrating how the declination of the Sun changes as Earth revolves around the Sun. A model solar system is very good for showing this, but the system is too small for a large class to see it clearly. Demonstrate the difference between the solar day and the sidereal day by having one of your students sit on a stool in front of the class to play the part of the Sun. You are the Earth, and the other students are stars. Stand in front of the student on the stool facing the class and explain to the others the part they are playing. Then revolve about the student on the stool, keeping your back to the student at all times. After one revolution has been completed, ask the person on the stool how many rotations you made. The answer will be “None,” because your back was in view at all times. Next ask the class (playing stars) how many rotations you made during one revolution. They will answer “One.” Repeat the demonstrations but make one rotation with respect to the student on the stool. This will be two rotations with respect to the class. They will get the point that there is always one more rotation to anyone outside of the rotating and revolving system.

Introduction to Physical Sciences 14e

Demonstrate how the solar day varies in length by having the lamp on the lecture desk play the part of the Sun; you play the part of the Earth. With a meter stick in your hand (this is more effective than just the extended arm), stand at the side of the desk facing the lamp (Sun). Hold the meter stick out toward the lamp so that, sighting over the meter stick, you observe the Sun. Next rotate counterclockwise and revolve counterclockwise (take a couple of steps) around the lecture desk and lamp. Rotate 360° with respect to your original position. Note that you cannot observe the sun by sighting across the meter stick but must rotate slightly more than 360° to see the Sun on your meridian. Repeat the demonstration. This time take three steps in revolving. Have the student’s note that you must rotate even more than last time because you have to move around farther in orbit. Thus the length of a solar day depends on the orbital velocity of Earth. The greater the orbital velocity, the longer the solar day, provided that the rotation time remains constant.

ANSWERS TO MATCHING QUESTIONS a. 10 b. 23 c. 22 d. 8 e. 3 f. 20 g. 15 h. 21 i. 1 j. 14 k. 9 l. 4 m. 16 n. 11 o. 5 p. 18 q. 12 r. 6 s. 19 t. 7

u. 13 v. 17 w. 2

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d 2. d 3. c 4. b 5. b 6. b 7. d 8. c 9. d 10. b 11. c 12. b 13. c 14. b 15. b 16. c 17. b 18. a 19. c

20. d

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. rectangular 2. origin 3. parallels 4. great circle 5. latitude 6. longitude 7. Coordinated Universal Time (UTC) 8. twenty-four 9. four 10. westward 11. altitude 12. zenith angle 13. 30°E 14. summer solstice 15. Cancer 16. seasons 17. zodiac 18. Sun 19. Common Era 20. top

ANSWERS TO SHORT-ANSWER QUESTIONS 1. The origin in a Cartesian coordinate system is defined as the intersection of the two perpendicular numbered lines that make up the coordinate system.

2. Longitude and latitude. 3. Answer is relative to student’s location and reference point.

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4. Latitude: 0° to 90°N and 0° to 90°S. Longitude: 0° to 180°E and 0° to 180°W. 5. The meridian running through Greenwich, England, which is taken as 0° longitude, and the equator, which is taken as 0° latitude. 6. (a) A meridian. (b) Parallels 7. The equator is defined as the reference point for latitude; therefore, the equator is 0° latitude, by definition. Since the equator is a parallel it does not have a longitude value (or you can say that it corresponds to all longitudes)

8. The center of the Earth. 9. All meridians are half circles, portions of a great circle. 10. All parallels are small circles except the equator, which is a great circle. 11. 75° for Eastern Time Zone, 90° for Central, 105° for Mountain, and 120° Pacific. 12. Ante meridian and post meridian. Greenwich Mean Time and Coordinated Universal Time. 13. The year is generally defined as the elapsed time needed for the Earth to make one complete orbit around the Sun. However, the year can be defined in different ways: The tropical year, the year of seasons, is the time between successive vernal equinoxes. The sidereal year is defined as the time it takes Earth to complete one orbit around the Sun (i.e. the orbit period of the Earth) with respect to a distant star.

14. To have more hours of sunlight when most people are awake, which also conserves energy. 15. A solar day is the elapsed time between two successive crossings of the same meridian by the Sun. A sidereal day is for a star other than the Sun. The solar day is longer than a sidereal day by about 4 minutes. 16. The time advances into the next day. 17. The rotation of the Earth on its axis. 18. A ships longitude can be measured by a marine chronometer, an accurate clock.19. The point directly over your head. The altitude and zenith angle make up the angles between the horizon and the zenith. 20. Estimating the position by dead reckoning. 21. (a) Between 23.5° N and the equator. (b) Between 23.5° S and the equator. 22. The Babylonian calendar is based on the motions of the Moon. The Gregorian calendar is an improved Julian calendar, both of which are based on the position of the Sun.

23. About June 21 and December 22. 24. (a) 23½° N; (b) 23½° S; (c) and (d) 0°. 25. The time for the Moon to go through its cycle of phases.

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26. Julian: adopted by Julius Caesar in 4 B.C. with introduction of leap year. Gregorian: adopted by Pope Gregory in 1582 to correct a 10 day discrepancy, along with the addition of a century leap year correction. 27. Every four years, except century years not divisible evenly by 400; that is, there are 97 leap years in every 400 years. 28. The lighted and dark portions of the Earth are equal in each hemisphere during the spring and fall equinoxes. During northern hemisphere summer the boundary of the circle of illumination moves such that the North Pole region receives 24 hours of daylight. The opposite happens in winter. 29. 2400 A.D. 30. The pole star changes over the precession period. 31. Earth’s precession is caused by gravitational forces (torques) from both the Sun and the Moon acting on the Earth. This occurs because Earth is not a perfect sphere.

32. 25,800 years.

ANSWERS TO VISUAL CONNECTION (a) North Pole, +90°; (b) parallel, +45°; (c) equator, 0°; (d) parallel, −60°; (e) South Pole, −90°; (f) meridian, 90° W

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Sundials measure apparent solar time. 2. The altitude of the North Star equals the latitude. 3. Yes, at the equator. All meridians are one-half of a great circle. The only parallel that is a great circle is the equator. 4. It is hotter in Texas because the sunlight is more direct (more intense). 5. On the summer solstice the Sun is 23.5° above the equator; therefore the Sun is 23.5° above your horizon on that day if you lived at the North Pole (the equator and the pole are 90° apart). Due to the rotation of the Earth, objects follow circular paths in the sky around the North Celestial Pole each day. Thus, the Sun would remain parallel to your horizon and move horizontally around the sky over 24 hours making a circular path around the sky. 6. (a) Yes. Parallels are complete circles. (b) No. Meridians are only half circles.

Introduction to Physical Sciences 14e

7. Born in a leap year, only a birthday every 4 years, and even less for a century year not divisible by 4 8. Precession of the Earth’s axis.

ANSWERS TO EXERCISES 1. Place A: 50° N, 90° W 90° N – 50° N = 40° • N.P. 90° N – 60° S = 150° Place B: 60° S, 90° E 150° + 40° = 190° 2. Place A: 20° N, 75° W 20° N – 0° = 20°

Place B: 30° S, 75° W 30° S – 0° S = 30° 20° + 30° = 50° 3. 190° × 60 n mi= 11,400 n mi 4. 50° × 60 n mi = 3000 n mi 5. 39° S. 103° E. 103° E is opposite 77° W. North Pole

6. 24° N, 133° E 7. Taking the central meridians of the time zones: 120° – 75° = 45°, or 3 hours, so 3 P.M., Oct. 8. 8. Taking the central meridians of the time zones: 105° – 75° = 30°, or 2 hours earlier, so 11 P.M., Oct 15 (crossed midnight). 9. Work the same way as you did Problem 7. Tokyo is in the 135°E time zone. The answer is 3 A.M. the next day—November 27.

Introduction to Physical Sciences 14e

10. Work the same way as you did Problem 7. The same figure can be used again. The answer is 3 A.M. the next day—February 23. 11. 5 a.m., MST 12. 1 a.m. 13. Drawing a time circle and counting from 105° W (Utah on MST) to 15° E, the time and date would be 3 A.M., Feb. 23. 14. Drawing a time circle and counting from 0° to 150° W, the time and date would be 4 A.M., July 1. 15. Drawing a latitude circle with zenith at 34° N and the Sun at 23.5° N, the zenith angle is 34° – 23.5° = 10.5°. So the altitude is 79.5°. June 21 the Sun’s declination is 23.5°N. 34°N – 23.5°N = 10.5° zenith angle Altitude = 90° – zenith angle = 90° – 10.5° = 79.5°. 16. 17.5° 17. June 21 the Sun’s declination is 23.5° N. 90° – 71.5° altitude = 18.5° zenith angle 23.5 °N + 18.5° = 42°N latitude 18. December 22 the Sun’s declination is 23.5°S. 90° – 65.5° altitude = 24.5° zenith angle. 24.5° north of 23.5°S is 1°N 19. On or about June 21, 74.5°.

θ = 39° – 23.5° = 15.5° 90° – 15.5° = 74.5° (maximum altitude) 20. On or about December 22, 27.5° 21. March 21 to June 21 = 92 days, and March 21 to April 15 = 25 days. Then, 23.5°/92 days (25 days) = 6.4° N. 22. 18.0° 23. L = ZA – Sun degrees = 50° N – 23.5° = 26.5° N 24. L = ZA + Sun degrees = 30° S + 0° = 30° S

Introduction to Physical Sciences 14e

25. L = ZA – Sun degrees = 55° – 15° = 40° S. (Sun overhead at 15° N on May 20. See Table 15.2.) 26. Dec. 14 to Dec 21 = 8 days (0.26°/day) = 2.1°, so Sun is overhead at 23.5° – 2.1° = 22.4° S. L = ZA – Sun degrees = 34° – 22.4° = 16.6° N 27. L = ZA – Sun degrees = 43.5° -23.5° = 20° S 5 h (15°/h) = 75° E 28. L = ZA – Sun degrees = 23.5 – 23.5° = 0° (equator) 10 h (15°/h) = 150° E

Introduction to Physical Sciences 14e

Chapter 16 THE SOLAR SYSTEM In this chapter, we essentially begin our study of astronomy as we move out into space. The chapter begins with a definition of astronomy followed by some historical considerations and Kepler’s laws. This sets the stage for comparing and contrasting the properties of the planets. Tables 16.1 and 16.2 are presented to summarize the properties of and comparisons between the terrestrial and Jovian planets. Dwarf planets are also defined and the few that have been discovered are highlighted. Also, the origin of the solar system is considered, along with the latest reported information concerning newly discovered planetary systems beyond our solar system. The two highlights explain how spacecraft travel to the outer solar system and how astronomers discover exoplanets. The conceptual questions and answers discuss how the Foucault pendulum works and how the rotations of gaseous planets are determined.

DEMONSTRATIONS Make a model solar system that shows motion of planets. Build a Foucault pendulum. Take care to use a long wire attached to a ceiling with a pivot. Use a large sphere to illustrate rotation and revolution of the Earth. Also use a large sphere to illustrate Eratosthenes' calculation. Show a class demonstration of parallax (finger of each hand held together, then one held close to the face and the other at arm's length).

ANSWERS TO MATCHING QUESTIONS a. 22

b. 6

c. 24 d. 8

e. 15

f. 4

g. 21

h. 10

i. 20

j. 7

k. 19 l. 13

m. 2

n. 11

o. 16 p. 5

q. 17

r. 1 s. 23

t. 9

u. 14

v. 18

w. 3 x. 12

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d 2. a 3. c 4. b 5. d 6. b 7. b 8. c 9. c 10. a 11. b 12. b 13. c 14. c 15. b 16. c 17. d 18. c

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. Astronomy 2. geocentric 3. major axis 4. orbit period 5. inferior 6. prograde 7. opposition 8. 0.3 9. Foucault pendulum 10. Mercury 11. greenhouse effect 12. iron oxide

Introduction to Physical Sciences 14e

13. Uranus 14. Ceres 15. Eris 16. condensation 17. transit or eclipse

ANSWERS TO SHORT ANSWER QUESTIONS 1. Planets, moons, asteroids, meteoroids, dwarf planets, comets, and dust. 2. A geocentric model is Earth-centered; a heliocentric model is Sun-centered 3. When farthest from the Sun— conservation of angular momentum. 4. T2 α R3 5. The major planets orbit in a flat disk with nearly circular orbits around the Sun. 6. Rocky planets near the Sun are placed in the terrestrial group, while the gaseous giant planets are in the jovian group. 7. No it’s a superior planet. 8. No, it is an oblate spheroid—bulging at the equator. 9. The Earth is about 4 times larger than the Moon. 10. Rotation: Foucault pendulum. Revolution: parallax and aberration of starlight 11. Mercury, Venus, the Earth, and Mars all have rocky surfaces, high densities, orbits close to the Sun, metallic cores. These characteristics resemble those of the Earth. 12. The Venusian surface is really hot (467°C or 873°F) due the large amount of carbon dioxide in the atmosphere which produces a large “greenhouse effect”. 13. Earth’s Grand Canyon is about 300 mi long and was carved from flowing water. Mars’ Vallis Marineris is a 3000 mi crack in the surface. 14. Terrestrial—solid (more dense) and small. Jovian—gaseous and large. 15. The rings of Saturn are clearly larger and more evident than the rings of the other jovian planets. Saturn’s rings are composed of ice and rock material which makes it much more reflective and easier to see in the sunlight. The rings of Jupiter, Uranus, and Neptune are much fainter due to the smaller amount of material they contain as well as the composition being dark, dusty material. Each set of rings shows individual bands and gaps due to the influence of gravity and the presence of shepherd moons. 16. Uranus’ axis lies almost in its orbital plane. 17. Methane in the atmosphere absorbs red light, reflecting blue light. 18. Dwarf planets are spherical objects orbiting the Sun that have not fully cleared their orbit path.

Introduction to Physical Sciences 14e

19. Pluto has not cleared its orbit (it crosses the orbit of Neptune) 20. A flatten region of dwarf planets and comets beyond the orbit of Neptune 21. The early cloud of dust and gas out of which the Sun and the planets formed. Liquids and solid material condensed out of this early cloud to form the planets. 22. Interstellar dust was key in allowing the condensation process to work. 23. Astrometry is the study of the precise positions of celestial objects. 24. The Doppler Effect and transits are methods to detect exoplanets.

ANSWERS TO VISUAL CONNECTION (a) Jupiter, (b) Mercury, (c) Sun, (d) asteroid belt, (e) Mars, (f) Saturn, (g) Uranus, (h) Earth, (i) Neptune, (j) Venus, (k) Pluto

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. The advance knowledge of metals, ceramics, rocket fuels, computers, communication, and the initiative of political leaders and the public to fund such research.. 2. This Foucault pendulum is swinging in a plane that is parallel to the Earth’s spin axis. The pendulum also swings in a plane fixed with a distant star, Polaris, the North Star in this case. As the Earth rotates the pendulum remains in the same plane, fixed with the north-south plane. Therefore, the pendulum does not rotate. 3. A complete set of phases are observed for Mercury and Venus because the angle between the planet-Sun-Earth viewing-angle changes significantly, between 0° and 180°. However, for the outer planets, the viewing angle does not change significantly and we see the outer planets in a near-full phase all the time. 4. Astronomers used hypotheses and tests in order to justify the changes to a planet definition. Reclassifying allows scientists to better compare properties of similar objects, learning more in the process. 5. The solar nebula theory explains why all the major planets orbit the Sun in a disk, in circular orbits, and all moving in the same direction. The dwarf planets are also explained because they lie in the same plane and move in the same direction. Rotations are not explained because some planets and dwarf planets have retrograde rotation.

Introduction to Physical Sciences 14e

ANSWERS TO EXERCISES 1. 𝑇 2 = 𝑘𝑅3 = 1.0

𝑦2 𝐴𝑈

5.22 1

= 141𝑦 2 𝑇 = √141𝑦 2 = 11.9 𝑦

2. 11,800 years 3

3. 𝑅3 = 𝑇 ⁄𝑘 =

0.6162 𝑦 2 = 0.38𝐴𝑈 3 1𝑦 2 ⁄𝐴𝑈 3

R = 0.724 AU

4. R = 30 AU 5. T = 2.8 y 6. T = 8 y 7. R = 2.8 AU 8. From Table 16.1, for Jupiter, RJ = 5.2 AU, and for Mars, RM = 1.5 AU. So, from Exercise 7 answer, closer to Mars. 9. T2 = kR3 Therefore, the smaller the R, then the smaller the T.. 10. T2 = kR3, so the smaller R means a smaller T. For a circle, orbit velocity is inversely proportional to time, and smaller time means greater velocity. 11. Mercury, Venus, Earth, Mars. 12. Jupiter, Saturn, Uranus, Neptune, Pluto.

Introduction to Physical Sciences 14e

Chapter 17 MOONS AND SMALL-SOLAR SYSTEM BODIES Students should be given an outside assignment to observe the Moon for one month. Have them determine the time of meridian crossing, and record the rising and setting times of the Moon. The instructor should point out how the frame of reference is formed for the surface of the Moon. Many astronomy texts show the Moon reversed left to right, because many telescope pictures are reversed. All photos in this text are oriented to show what someone on the Earth would see if looking at the Moon with the unaided eye.

DEMONSTRATIONS Use a model solar system with a Moon that shows the motion of the Earth and the Moon and also a large Earth sphere and an 8-in.-diameter Moon that the instructor can move manually around the Earth sphere. Let the classroom wall be the Sun. Explain to the class the position of the Moon at new, first quarter, full, and last quarter phases. With the Moon at any of these positions, rotate the Earth to show the rising and setting times of the Sun and the Moon. Demonstrate with a model sphere of the Earth and the Moon, plus an incandescent lamp for the Sun, the positions necessary for solar and lunar eclipses.

ANSWERS TO MATCHING QUESTIONS a. 12

b. 2 c. 19 d. 23 e. 24 f. 22 g. 17 h. 5 i. 4

j. 1 k. 13 l. 18 m. 14

n. 6

o. 8 p. 25 q. 10 r. 20 s. 7 t. 16 u. 15 v. 9 w. 21 x. 11 y. 3

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b 2. c 3. b 4. b 5. a 6. c 7. c 8. b 9. d 10. a 11. a 12. c 13. b 14. c 15. a 16. d 17. b 18. d 19. d 20. c 21. c 22. a 23. a 24. b 25. c 26. e

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. highlands 2. rays 3. later 4. 6 A.M. 5. full 6. waning 7. lunar 8. umbra 9. spring 10. twelve 11. zero 12. asteroids 13. Ganymede 14. Io 15. Titan 16. Charon 17. Water ice 18. Asteroids 19. meteorite 20. nucleus

Introduction to Physical Sciences 14e

ANSWERS TO SHORT-ANSWER QUESTIONS 1. Craters, maria, rays, rills, mountains, basins, faults. 2. The lunar mare are lower elevation regions on the Moon, probably formed by huge impacts early in the Moon’s formation. These lowlands were later filled with lava which is darkercolored rock 3. (July 20) 1969. There were five other moon landings. 4. 3.1 – 4.4 billion years. 5. 29.5 days for both. 6. Because the Moon gains 12.2°/day on the Sun. 7. Waxing: illuminated portion of the Moon growing or increasing. Waning: illuminated portion of the Moon decreasing. 8. Crescent: less than one quarter of the surface illuminated. Gibbous: more than one quarter of the surface illuminated. 9. An orbit in which the rotation period equals the revolution period. 10. Umbra: region of total darkness of the Moon’s shadow. Penumbra: semidark region of the Moon’s shadow. 11. First quarter, Sun to west of Moon. Last quarter, Sun to east of Moon. 12. A partial eclipse of the Sun occurs when the Moon moves in between the Earth and the Sun, but alignment is not perfect. The penumbral (semidark) shadow is cast on the observer and only part of the Sun is covered. 13. Moon is not in the Earth’s complete shadow. 14. Lunar: Sun, Earth, and Moon. Solar: Sun, Moon, and Earth. 15. Gravitational attraction different on opposite sides. The Earth is “pulled away from water. 16. Spring tides: at new or full phases, greatest variation. Neap tides: at first or last-quarter phases, minimum variation. 17. Three, the Earth 1 and Mars 2. 18. Two: Phobos and Deimos. 19. Four. These four largest moons on the planet Jupiter were discovered by Galileo. 20, Jupiter’s Io, caused by Jupiter’s gravity. 21. Ganymede, Jupiter.

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22. Saturn’s large moon Titan is unique because it has a thick hydrocarbon atmosphere, mostly methane gases. The hydrocarbon smog may have some interesting chemistry going on. The surface of Titan also shows channels and lakes of liquid methane. Evidence also suggests it might have a liquid subsurface ocean. 23 Neptune’s moon Triton is interesting because it has geologic activity in the form of geysers spewing out of its icy surface. The surface shows complex landforms and poles of frozen nitrogen. Triton also orbits retrograde (backwards) and thus is believed to be a gravitationally captured moon. 24. Titan, Saturn; Titania, Uranus; Triton, Neptune. 25. Rotation and revolution are the same (like the Moon and the Earth) 26. Hydra 27. Eight times smaller 28. Dwarf planets are probably composed primarily of water ices mixed with some rocky material. Because of their great distance from the Sun, the composition is similar to comets. 29. Most asteroids are between the orbits of Mars and Jupiter. 30. A meteor when it enters the Earth’s atmosphere, and a meteorite if it strikes the Earth’s surface. 31. Comet tails are composed of two different materials: gas and dust. The gases are charged particles of hydrogen, oxygen, nitrogen, and carbon. The dust particles are very small grains of carbon-type particles. 32. 76 years. 33. Sunlight reflected from dust particles.

ANSWERS TO VISUAL CONNECTION (a) new, (b) waning crescent, (c) 3rd quarter, (d) waning gibbous, (e) full, (f) waxing gibbous, (g) 1st quarter, (h) waxing crescent

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Yes. You will observe phases of Earth similar to the phases we observe of the Moon with the new phase beginning when the Moon, Earth, and the Sun are in the same plane with Earth between the Sun and the Moon.

Introduction to Physical Sciences 14e

2. The phases of the Moon would essentially occur in the opposite order if the Moon orbited backwards. Following the new Moon, would be the crescent, then the 3rd quarter, gibbous, full, gibbous, 1st quarter, crescent, then back to new Moon. But the 3rd quarter moon would have to be renamed to 1st quarter since it comes first. The appearance of the phases would not change, but the direction the Moon moves in the sky would be westward instead of eastward. 3. The crescent Moon is pointing the wrong direction. It appears the sunlight is coming from above. 4. The angular sizes are the same because the Sun is about 400 times larger than the Moon, but the Sun is about 400 times more distant. 5. Charon would go through all phases, but it would always be seen in the sky from Pluto because of the synchronus orbit.

ANSWERS TO EXERCISES 1. Approximately 133 N. 2. Approximately 27 1b. 3. Approximately 354 days (29.5 × 12). 4. Approximately 328 days (27.5 × 12). 5. 6:50 AM 6. The Moon rises about 50 minutes later each successive day. Seven days later the Moon will rise about 6 hours later. For example, the new Moon rises at 6:00 am; after 1 week it is a 1st Quarter Moon which rises at 12 noon. 7. (a) The right side or west side is bright, because the Sun has just set in the west. (b) It would be the same phase in Australia and the west side would be bright, but this is the left side. 8. (a) The eastern side or the left side. (b) The same phase (last quarter), and the east or right side would be bright. 9. Halfway through waxing crescent phase is 1/8 cycle. Last quarter phase is 3/4, or 6/8, cycle. So, 6/8 – 1/8 = 5/8, and 29.5 days (5/8) = 18 days. 10. From the start of a waning gibbous phase at 12 midnight to the end of the waning crescent phase at 12 noon is 1/2 cycle. Then, 29.5 (1/2) = 14.8 days.

Introduction to Physical Sciences 14e

11.

12.

13. (a) High tides are approximately 180° apart. 180° from 84°W is 96°E. (b) Longitude 90° east and west of 96°E is 6°E and 174°W. 14. (a) 180° from 95°W is 85°E. (b) Longitude 90° east and west of 85°E is 5°W and 175°E.

Introduction to Physical Sciences 14e

Chapter 18 THE UNIVERSE The universe as viewed from planet Earth is composed of stars in the Milky Way galaxy and other galaxies. The galaxies, in turn, form the myriad of galaxy clusters that are observed throughout the depths of space. This chapter discusses these basic building blocks (stars and galaxies) and presents information on the structure, shape, form, age, and extent of the known universe. The origin of the universe is slowly being understood, and the basic ideas behind our understanding are presented in this chapter. The beginning of the chapter is devoted to the concept of the celestial sphere and celestial coordinates. After a thorough overview of our star (the Sun), other stars are classified and compared. The birth, life, and death of stars are fairly well understood, and the discussion includes low and high mass stars and their fates (white dwarfs, neutron stars, and black holes).The Hertzsprung-Russell diagram should be explained and made familiar to the student. A take-home open-book quiz is an excellent way for the student to learn how the diagram is plotted and what it explains. This chapter also presents information on the Milky Way galaxy, other galaxies, the expansion of the universe, dark energy, and dark matter. The latest information on the modification of the standard model of the Big Bang is also discussed. Highlights of how astronomers determine distances and the age of the universe are included. Three conceptual questions and answers help students overcome misconceptions in astronomy.

DEMONSTRATIONS Use a transparent globe to show star clusters and the motion of the Earth and the Sun with respect to the stars. Color slides showing the spectrum of a nearby star and a star moving rapidly away from the Earth are helpful in illustrating the Doppler shift. A variety of pictures of celestial objects is available on the Internet. Models of the galaxy can also be drawn and cut out on cardboard; calculations of the overall size and scale are a good way to give perspective. A good demonstration of the expansion of the universe is made by grid lines drawn onto a transparency, then enlarge this by 20% on a photocopy machine. Make another enlargement of 40% onto

Introduction to Physical Sciences 14e

another transparency. Overlay the transparencies to demonstrate how the spatial grid (the universe) expands.

ANSWERS TO MATCHING QUESTIONS a. 3 b. 21 c. 1

d. 13 e. 18 f. 25 g. 6 h. 15

n. 22 o. 27 p. 4 q. 16

r. 20 s. 23 t. 2

u. 7

i. 11 j. 17 k. 12 v. 24

w. 14

l. 5

x. 10

m. 8 y. 26

z. 19

aa. 9

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. a 2. a 3. d 4. b 5. d 6. a 7. a 8. b 9. d 10. d 11. b 12. a 13. b 14. b 15. a 16. d 17. b 18. d 19. c 20 a

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. parallax 2. celestial equator 3. right ascension 4. declination 6. helium 7. Flares 8. constellations 9. blue 10. spectrum 11. brightness 12. main sequence 13. Red giants 14. contract 15. Nebula 16. nova 17. neutron star 18. supernova 19. supermassive 20. 13.7

ANSWERS TO SHORT-ANSWER QUESTIONS 1. Cosmology is the study of the structure and evolution of the universe. 2. Declination, right ascension, distance. 3. The ecliptic and celestial equator are tilted by 23.5°. The reason for this is the 23.5° tilt of the Earth’s axis. 4. The vernal equinox is the point of intersection of the celestial equator and the ecliptic, where the Sun is moving northward. The celestial prime meridian is the great circle passing from the North Celestial Pole to the South Celestial Pole perpendicular to the celestial equator and intersecting the celestial equator at the vernal equinox. 5. The parsec is larger. One parsec equals 3.26 1y. 6. Stellar parallax is caused by the motion of the Earth about the Sun. In order to calculate distances to a star, astronomers measure the parallax angle of the star, and, using the known Earth-Sun distance, triangulation is used to solve the distance to the star. 7. 5800 K.

Introduction to Physical Sciences 14e

8. Stellar parallax is caused by the motion of the Earth about the Sun. In order to calculate distances to a star, astronomers measure the parallax angle of the star, and, using the known Earth-Sun distance, triangulation is used to solve the distance to the star. 9. The Sun’s corona is visible only during a solar eclipse. 10. The Sun is approximately 100 times bigger than Earth. 11. The core, the radiation and convection layers, the photosphere (“surface”), the chromospheres, and the corona. 12. 4 1H1 → 2He4 + 2(0e+1 + ν + γ) + energy 13. A star is a giant ball of gas which is stable because of the balance of two forces. Pressure pushes the gas outward, but gravity pulls the gas inward. 14. Apparent magnitude is a measure of the brightness of a star as observed from the Earth. Absolute magnitude is a measure of the brightness a star would have if it were positioned 10 parsecs from the Earth. 15. Over one day, the Moon, the stars, and the constellations move across the sky from East to West. Because Earth rotates eastward, the objects in the heavens will move westward, rising in the East and setting in the West. 16. Proxima Centauri is 4.3 ly distant. 17. O, B, A, F, G, K, M. 18. A main sequence star is Sirius, a red giant is Arcturus, a supergiant is Betegeuse, and a white dwarf is Procyon B. 19. An emission nebula is a bright nebula in which hydrogen gas is ionized by the energy from nearby stars and glows by fluorescence. A reflection nebula is a bright nebula in which the energy of nearby stars us merely reflected and scattered. 20. The Sun will become 100 times larger as a red giant. 21. Nucleosynthesis is the creation of nuclei inside stars. 22. The mass of a star determines its lifetime. The mass determines how big and hot the star will get during its life. The temperature is very important because the temperature determines how quickly the nuclear reactions take place in the core of the star. The faster these nuclear reactions take place, the shorter the lifetime of the star. 23. A brown dwarf is a failed star, one unable to sustain fusion long enough to move onto the main sequence. About 100 brown dwarfs have been identified.

Introduction to Physical Sciences 14e

24. Protostar, main-sequence star, red giant, planetary nebula, white dwarf. 25. A supernova is a much more powerful explosion than a nova. 26. A white dwarf object is about 100 times smaller than the Sun (same size as Earth). A neutron star is about 1300 times smaller than a white dwarf, or 130,000 times smaller than the Sun. 27. A pulsar is a rapidly rotating neutron star. It emits energy along its magnetic poles that are spinning around in space. When that pole is pointing directly toward Earth, we receive a burst of emission; the emission turns on and off as the pulsar rotates. This is similar to a lighthouse because the light in a lighthouse turns “on” and “off” as it rotates around. 28. (The sketch should resemble Fig. 18.22.) 29. A black hole does not emit energy so it cannot be “seen”. However, astronomers can detect the presence of a black hole by its gravitational influence on nearby objects. Typically the black hole is in a binary system so astronomers can observe the motion of the companion and infer the presence of the unseen object. 30. Spiral, elliptical, irregular. 31. Quasars are the cores of galaxies that were forming when the universe was young. They are the most distant objects observed. 32. The two types of spiral galaxy are the normal and barred type galaxies. The normal types have a round nuclear region with spiral arms that curve out from the nucleus. The barred spiral galaxies have an elongated nucleus (shaped like a rectangular bar) with the spiral arms curving out from the ends of the nuclear bar. 33. Dark matter is unobserved matter hypothesized by astronomers to explain why a cluster of galaxies exists as a gravitationally bound system. Dark energy seems to be accelerating the expansion of the universe. 34. The Milky Way is a normal spiral galaxy about 100,000 ly in diameter. Its disk is about 2000 ly thick, and its nuclear bulge is about 20,000 ly thick. 35. In a halo around the Milky Way lie about 150 globular clusters, which are spherical collections of a few tens of thousands of stars. Their distribution led to the conclusion that our solar system was located in a spiral arm of the Galaxy, not at its center. 36. Dark matter is one of the greatest mysteries in astronomy. The composition of dark matter is unknown at this time, but it is believed to exist in two forms: ordinary matter and exotic matter. The ordinary matter is believed to be small, dense objects that we cannot see because they are

Introduction to Physical Sciences 14e

too dim. The exotic matter is believed to be subatomic particles that we have and have not discovered. 37. Astronomers classify the Milky Way as a barred spiral galaxy, type b (SBb). 38. Hubble’s law states that the greater the recessional velocity of a galaxy, the farther away the galaxy. Hubble’s law indicates that the universe is expanding. 39. Galaxies are not evenly distributed in space but congregate in huge superclusters and long filaments that are separated from each other by huge voids containing very few galaxies. 40. The cosmological redshift, the cosmic microwave background, and the H/He ratio. 41. The most recent modification to the Big Bang model is called the inflation model. It differs from the standard model in that it includes a tremendous release of energy at 10-35 seconds after the big bang. This tremendous release of energy caused the universe to expand (inflate) very rapidly and thus flatten space-time. The release of energy is thought to be caused by the breakup of a unified force as the universe cooled slightly from the beginning. 42. The universe is about 71.4% dark energy, 24% dark matter, and 4.6% normal matter.

ANSWERS TO VISUAL CONNECTION (a) Luminosity (LSun); (b) Temperature (K); (c) White Dwarfs (hotter, dim, small, blue-white color); (d) Main Sequence (upper part: hotter, blue, bright, massive; middle part: Sun-like; lower part: coller, red, dim, least massive); (e) Giants (cooler, bright, red); (f) Supergiants (cooler, bright, red, largest size)

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Astronomers took a long time to accurately measure the distances to stars because the distances are so far away. It took more modern technology and telescopes to provide precise enough measurements. Note however that the basic method to measure stellar distances was known very well by the Greeks over 2000 years ago. 2. If Sirius exploded into a supernova it would be a cataclysmic event for anything within about 50-100 light-years distant. The Crab supernova released so much energy from about 6000 lightyears from Earth that it was brighter than the full Moon for two weeks. Sirius is only 8.7 lightyears from Earth; it would release so much energy in the form of gamma rays and x-rays that it would probably destroy all life on the Earth.

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3. Star A is blue, star B is yellowish, and star C is reddish. Because of the Stefan-Boltzmann law (ET4), higher temperature stars emit more light. Star A is the brightest star. 4. Follow a line through the stars Merak and Dubhe in the “bowl” of the Big Dipper. Look for an isolated, magnitude 2 star (Polaris). See Appendix IX. 5. Like the Sun, emission nebulae produce their own light. Like the Moon, reflection nebulae just reflect light given off by stars. 6. It will be several billion years before the Sun reaches the red giant phase. 7. The future of the universe depends on the overall energy density. If the density is over a critical value then the universe will stop expanding and collapse. If the density is the critical value then the universe will continue to expand forever, but take an infinite amount of time to do so. If the universe is less than the critical value then the universe expands forever. With the recent discovery of the dark energy accelerating the expansion of the universe, the estimated future of the universe is a faster expansion over time.

ANSWERS TO EXERCISES 1. d = 1/p = 1/0.20 arcsec = 5.0 pc 2 d = 1/p = 1/0.38 arcsec = 2.6 pc (=8.6 ly) 3. (365 d/y)(24 h/d)(60 min/h)(60 s/min) = 3.15 × 107 s/y 4. (1.86 × 105 mi/s)(3.15 × 107 s/y) = 5.86 × 1012 mi/y 5. 196 pc Use d=1/p, where distance, d is in parsecs and the parallax angle, p is in arcseconds. Thus p =1/130 pc = 0.008 arcseconds. 6. (130 pc)(3.26 ly/pc) = 420 ly 7. –3 is 10 magnitudes brighter than +7, and so it is 100 × 100 = 10,000 times brighter. 8. Venus is 5 magnitudes brighter than the star, so it is 100 times brighter 9. 211 Mpc Use Hubble’s Law Vr = Hd, where H = 73 km/s/Mpc. The distance to the galaxy cluster can be determined from d = Vr/H = 15,000 / 73 = 205 Mpc (mega-parsecs). 10. For the Hydra supercluster having a distance of 800 Mpc, the recessional velocity is calculated from Vr = Hd, Vr = 73*800 = 58,400 km/s.

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100

11. The age of the universe is calculated from Hubble’s constant, the rate of expansion of the universe. If H = 100 km/s/Mpc then, Age (in billions of years) = (1 x 1012)/H = (1 x 1012)/100 = 1 x 1010 years = 10 billion years. If H = 50 km/s/Mpc then, Age (in billions of years) = (1 x 1012)/50 = 2 x 1010 years = 20 billion years. 12. If H = 75 km/s/Mpc, the age of the universe is Age (in billions of years) = (1 x 1012)/75 = 1.33 x 1010 years = 13.3 billion years.

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101

Chapter 19

THE ATMOSPHERE For students to understand and appreciate weather principles and phenomena, it is essential that they have a knowledge of the fundamental physical properties of the atmosphere. The current emphasis on the environment offers a timely introduction to the study of the atmosphere. Everyone is aware of atmospheric pollution problems (Chapter 20); however, to really understand, students should know the normal conditions of our atmosphere that are being jeopardized. These conditions are presented in this chapter. The chapter discusses the composition of the atmosphere and how the relative percentages of the major constituents are maintained. Attention is given to the physical properties that distinguish one part of the atmosphere from another, particularly the divisions based on temperature. The section on the energy content of the atmosphere gives the student an understanding of what is responsible for the dynamics of the atmosphere. As pointed out in Chapter 1, measurement is necessary to describe conditions, and this also applies to the atmosphere. Common atmospheric measurements, with which the student should be familiar, are discussed. The concept of relative humidity and the dew point temperature should be covered thoroughly. The last two sections of the chapter are concerned with the movements of the gases of the atmosphere-winds and air currents-and the associated formation of clouds. Convection cycles and their relationship to local winds and world circulation patterns have important effects to which the student can relate. This is what makes the study of meteorology so interesting. Clouds are one of the most common sights in a student’s environment. This promotes interest in the study of various cloud types and characteristics. A knowledge of cloud formation is prerequisite to understanding precipitation processes, mechanisms, and types, which will be considered in the next chapter.

DEMONSTRATIONS 1. Atmospheric Pressure A dramatic demonstration of atmospheric pressure is the crushing of a metal can. There are two variations of this demonstration. One way is to use a gallon metal can (e.g., a well-rinsed paint thinner can). Place a small amount of water in the can and heat the open can sitting on a ring stand with a Bunsen burner. When the water is boiling vigorously, as evidenced by condensed

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water vapor coming from the can (not steam, which is invisible), turn off the burner and place a rubber stopper securely in the can opening. (A rubber stopper is recommended over a screw, gasket cap. The rubber stopper is safer should you forget to turn off the burner.) As the can cools and the steam condenses, a partial vacuum forms and the pressure difference causes the can to slowly collapse. After the can is crushed, slowly remove the stopper. An audible rush of air illustrates the partial vacuum in the can (like that in a vacuum-packed coffee can). A quicker, and perhaps more convenient, way to demonstrate atmospheric pressure is to use an aluminum soft drink can. (Large metal cans are sometimes difficult to obtain.) Place a small amount of water in the rinsed can, and hold the can over the flame of a Bunsen burner using a pair of tongs. Upon hearing vigorous boiling, quickly invert the open top of the can into a container of water. The can is crushed immediately. When the can is raised from the water, water will run from the can, having been forced into the partial vacuum. (Have an extra can available, because students will want to see the demonstration repeated.) The hole of a small can may also be stopped with a piece of clay. 2. Dew Point and Atmospheric Pressure For a classroom demonstration of dew point and atmospheric pressure, an iced soft drink in a glass with a straw can be used. After studying the section on humidity, the student should be able to explain why condensation occurs on the glass. This demonstrates that the air in the vicinity of the glass has been cooled to its dew point, and because the air is saturated (100% relative humidity), the water vapor condenses on the glass. Drinking through a straw demonstrates atmospheric pressure. Aspiration through the straw reduces the pressure within, and liquid is forced up the straw by the pressure difference relative to the outside atmospheric pressure. This can be effectively shown by punching a small hole in the side of the straw. This equalizes the pressure and makes the straw useless. 3. Humidity A simple psychrometer can be easily constructed. The only essential equipment is two thermometers, a water reservoir, and a cloth wick to keep one of the bulbs wet. Care should be taken not to place the wet bulb too near the water reservoir, as this will impair free evaporation. One method is to use a paper quart milk container as a reservoir and attach the thermometers to the side using garbage bag ties. A hole can be made for the wick to extend into the water. The relative humidity and the dewpoint temperature can be determined from the difference in the thermometer readings by using the tables given in the text Appendix VIII. This demonstration can be set up during one class period and a reading taken the next, to allow time

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for the evaporation rate to come to equilibrium. The students themselves can construct a psychrometer as a project, taking readings in their dormitory for a period of days and plotting the results to show the humidity variation over time. 4. Weather Instruments Weather-measuring instruments, such as barometers and anemometers, can be used to demonstrate how measurements are obtained. See the companies listed in the Teaching Aids section. 5. Convection Cycles A commercially available product called a Lava-Lite works well in demonstrating convection cycles. The Lava-Lite consists of a lamp base with an upper portion containing liquids of different densities. One liquid is colored and rises in the other as a result of heat from the lamp. As the globules of liquid rise and cool, they fall back to the bottom of the container. The LavaLite should be turned on an hour or more before the scheduled demonstration, as time is required to initiate the action.

ANSWERS TO MATCHING QUESTIONS a. 7 b. 14 c. 18

d. 22 e. 3 f. 29 g. 13

n. 6

o. 25 p. 16

bb. 26

cc. 12

q. 9 r. 17

h. 24 i. 11

j. 20 k. 2

u. 28 v. 10

w. 27

s. 21 t. 1

l. 30 m. 15

x. 4 y. 23

z. 19 aa. 8

dd. 5

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. a

2. c

3. a

4. b

5. a

6. d

7. d

8. a

9. a

10. d 11. d 12. b

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. Meteorology 2. oxygen 3. stratosphere

4. carbon dioxide, CO2

7. maximum

10. land

8. opposite 9. sea breeze

5. 14.7

11. counterclockwise

6. 76 cm, 30 in. 12. low

ANSWERS TO SHORT-ANSWER QUESTIONS 1. 78% nitrogen, 21% oxygen, and 0.9% argon. 2. No, photosynthesis by plants replenishes the oxygen. 3. Chlorophyll. 4. (a) Decreases.

(b) Increases.

(d) Increases.

5. It absorbs and protects us from harmful ultraviolet radiation in sunlight.

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6. Solar disturbances, and the recombination of ionized gas molecules and electrons in the Earth’s upper atmosphere. 7. That it reflects 33% of incident sunlight. Much larger than the Moon, which has an albedo of 7%. 8. (1) Absorption of terrestrial radiation, (2) latent heat of condensation, (3) conduction from Earth’s surface. 9. Extreme scattering through dense atmosphere mixes colors-- whilte 10. Least scattering, and so can be seen for the greatest distance. 11. No, in actual greenhouses, glass is an important factor. 12. A change in temperature shifts the wavelength of the radiation such that it is transmitted or absorbed, and the Earth cools and heats periodically. 13. Temperature, thermometer; pressure, barometer; humidity, psychometer; wind direction and speed , anemometer. (Also, precipitation, rain gauge.) 14. Pressure on a liquid pool to support a column. 76 cm (30 in.). 15. The temperature of the air in the vicinity of the glass is lowered to the dew point. 16. The relative humidity is 100% at height where the rain is formed. 17. In the direction from which the wind is coming, because of the fins. 18. Wind fills the pivoted sock, and the tail of the sock points in the direction toward which the wind is blowing. The angle of the sock relative to the horizontal is an indication of the wind speed. (With a high wind speed, it stands "straight out.") 19. Doppler radar can give wind speed and direction, whereas conventional radar cannot. 20. A thermal cycle set up by localized heating. Near a body of water, land breezes and sea breezes are set up during the night and day, respectively.. 21. As viewed from above, in the Northern Hemisphere, cyclones rotate counterclockwise and anticyclones rotate clockwise. The rotations are opposite in the Southern Hemisphere. 22. Because of being in the westerlies wind zone. 23. West, upwind. 24. Windward side may need extra insulation. 25. (a) Low.

(b) High.

26. (a) Cirrocumulus.

(d) Low. (e) Vertical development. (c) Cumulonimbus.

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27. The rate of temperature decrease with altitude. Condensation takes place when the dew point of rising air is reached. Greater the lapse rate, the greater the cooling.

28. Most clouds appear white. This is because the water droplets scatter all of the visible wavelengths (colors) of the sunlight. Together the colors make up white light.. Clouds may be dark or gray when either a cloud is in another’s shadow or the top of a cloud casts a shadow on its own base. Nimbus clouds with dense, larger droplets appear dark.

ANSWERS TO VISUAL CONNECTION a. cumulonimbus g. altocumulus

b. cumulus

h. stratus

c. cirrus

d. cirrocumulus

i. stratoculumus

e. cirrostratus

j. nimbostratus

k. advection

f. altostratus l. radiation

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. (a) Lower specific heat, less heat to lose. Also, not mixed as in water. (b) Very dry and little water vapor or clouds to absorb terrestrial radiation and insulate from heat loss. 2. Summer monsoon – sea breeze. Winter monsoon – land breeze 3. Rain comes from cumulonimbus clouds that occupy a particular vertical area. 4. It would appear to be deflected to the left as westward points rotate underneath. 5. The pressure above the water in the cup would be reduced as the water drains and stream would eventually stop. Outside pressure equals inside pressure.

ANSWERS TO EXERCISES 1. (a) (50 – 16 km)/16 km = 2.1

(b) (80 – 50 km)/16 km = 1.9

2. Graph. 3. T = To – Rh = 70°F – (3.5F°/1000 ft)(14,000 ft) = 21°F (where R is rate). 4. T = To – Rh = 20°C – (6.5°C/km)(10 km) = – 45° C 5. From tables: (a) 46%

(b) 62°F (c) 12.7 gr/ft3 3

(d) AC = MC x RH = (12.7 gr/ft V )(0.40) = 5.8 gr/ft3 6. From tables: (a) 65%

(b) 76°F (c) 14.8 gr/ft3

(d) AC= (14.8 gr/ft)(0.65) = 9.6 gr/ft3

7. (a) AC = MC x RH = (23.4 gr/ft3)(0.90) = 21 gr/ft3 (b) DP= 101°F (table), and 105°F – 101°F = 4°F 8. (a) AC = (2.4 gr/ft3)(0.45) = 1.1 gr/ft3

(b) No. Coolness is due to water evaporating (latent

heat removed from bulb). Air temperature is above freezing, and water is replaced from reservoir (also at 35°F).

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106

(b) DP = 59°F (from table).

10.  T = 5°F (from Table VIII.1), and DP = 73°F (from Table VIII.2), so 80°F – 73°F = 7°F 11. No, the temperature and cloud formation would be 60°F. 12. 70° – 60° = 10°F

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Chapter 20

ATMOSPHERIC EFFECTS Having covered the fundamentals in the preceding chapter, we now turn our attention to the processes and dynamics of weather phenomena. Since we have studied cloud formation, we can now look at precipitation processes and types, storms, and other weather phenomena. In general, the movement of large air masses across the country affects our weather changes. A person who knows about the movements and characteristics of these air masses is better prepared to understand and predict changes in the weather. In this chapter, students will encounter terms familiar to them from weather forecasts, such as fronts and air masses. The material is easily related to daily observations, and particularly at this point, students should be encouraged to understand how meteorological conditions affect the environment. A section is devoted to sensational weather phenomena—storms. These atmospheric disturbances sometimes give rise to property damage and even death. The properties and characteristics of some of the most powerful storms are discussed. Also, safety procedures are emphasized, particularly for tornadoes and hurricanes. A Highlight discusses El Niño and La Niña, whose oceanic effects on the atmosphere have received a great deal of attention in the past decade. The environmental and climatic effects of air pollution are major concerns today. Discussions of these effects commonly appear in daily newspapers and can be heard on radio and TV. Having studied the atmosphere, the student can now better understand and appreciate these problems. A brief history of air pollution is given, and the chief pollutants are identified, along with their sources. Discussions such as the one on acid rain point out how pollution affects our environment. Also, the ramifications of ozone depletion and the ozone hole over the South Pole are discussed. The long-term climatic effects of atmospheric pollution, for which we are responsible, can only be speculative. However, there is much concern over the emission of greenhouse gases and global warming

DEMONSTRATIONS Chemical reactions of some air pollutants, SO2 for example, can be demonstrated in the classroom. However, the utmost care should be taken, and this type of demonstration is not advised. A safer demonstration is the collection of particulate matter by using filter paper. A damp filter paper can be left exposed in the classroom, or large volumes of air can be filtered for

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particulates by using a vacuum cleaner. The particulate matter on the filter may then examined with a scanning microscope. Film slides may be made if such equipment is available. As a class project, students can collect samples at different outside locations. Commercial air-sampling kits are also commercially available.

ANSWERS TO MATCHING QUESTIONS a. 9 b. 23 c. 10 n. 1

d. 17 e. 4

o. 14 p. 22

f. 20

q. 6 r. 24

g. 12

h. 3

s. 21 t. 13

i. 16

j. 7

u. 2 v. 15

k. 25

l. 11

m. 19

w. 8 x. 18 y. 5

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b

2. a

3. d

4. d

5. c

6. c 7. b

8. d

9. a 10. b

11. d 12. c

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. coalescence 8. warning

2. rain

9. 24

3. temperature

10. subsidence

4. cP

5. warm 6. occlusion

7. heat

11. photochemical 12. CFCs

ANSWERS TO SHORT-ANSWER QUESTIONS 1. Bergeron process essentials: (1) ice crystals, (2) supercooled water vapor, and (3) mixing. Silver iodide crystals are substituted for ice crystals, and dry ice is used to cool vapor and form ice crystals. 2. (a) No. Frost is the deposition of water vapor.

(b) Successive cycles and condensation

in cumulus clouds. 3. Classified according to temperature and moisture content. Air masses acquire characteristics of their source region. 4. See Table 20.1 and Figure 20.4. 5. The boundary of air masses. Warm occluded

cold

stationary

, and

6. See Section 20.2 for descriptions. The sharpness of a front's vertical boundary gives an indication of the rate of change of the weather. In general, cold fronts have sharper vertical boundaries than warm fronts, and hence lead to more sudden weather changes.

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7. (a) Within a cloud (intracloud discharge), between clouds (cloud to cloud discharge), and cloud to ground (ground discharge).

(b) Heat lightning: below horizon or

behind clouds. Bolt from the blue: lightning in clear air. 8. Resuscitation and keep warm (shock). 9. Warm front. A warm front advances over colder air. If the temperature of the cold air and Earth’s surface is below freezing, precipitation falling may cool and freeze on contact, producing an ice storm. 10. The tornado. Although the hurricane has more energy, the energy of a tornado is concentrated in a small region, giving a greater energy density. The larger hurricane, however, is usually more destructive on making landfall. 11. As described in Section 20.3 topic on Tornado Safety. 12. Latent heat. When the wind speed reaches 74 mi/h (118 km/h). 13. A hurricane watch is issued for coastal areas when there is a threat of a hurricane within 24 to 36 hours. A hurricane warning indicates that hurricane conditions are expected within 24 hours. 14. (a) August-September, for the North Atlantic. (b) Varies from state to state. Generally, April-August. 15. Any atypical contribution to the atmosphere resulting from human activities. 16. No. England had air pollution in the late 1200s. 17. Radiation and subsidence temperature inversions, traps pollution near ground. 18. Complete—CO2 and H2O; incomplete—CO, hydrocarbons, and soot (carbon). 19.. The high temperatures of complete combustion cause a reaction between the nitrogen and oxygen of the air. Contributes to acid rain and photochemical smog. 20. Classical smog is smoke-fog. Photochemical smog results from photochemical reactions of hydrocarbons and other pollutants with oxygen in the presence of sunlight. Ozone is the prime indicator of photochemical smog. 21. Rising air meets colder air and spreads out 22. Sulfur.

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23. Sulfur dioxide (and nitrogen oxides) combine with water to form acids, which fall as rain (or other types of precipitation). The acid raises the pH of bodies of water. The problem is most acute in the northeastern United States and eastern Canada because of major industrial areas to the west, but acid precipitation may now be found almost everywhere. 24. (a) Transportation. (b) Stationary sources (for example, electrical generating plants). (c) Industry. (d) Transportation.

(e) Photochemical smog.

25. Long term, average weather conditions. 26. (a) Increase in the greenhouse effect and global warming, (b) decreasing transparency of the atmosphere to isolation, (c) particulate matter and gaseous chemical reactions in the stratosphere. 27. Increased ultraviolet radiation and possibility of skin cancer. 28. The interaction of CFCs with the ozone layer could deplete the ozone and allow more UV to reach Earth, thereby increasing Earth's temperature. In the extreme, an increase in the average temperature could affect the environment (e.g., lengthen the growing season) and melt the polar ice caps. 29. There are no natural mechanisms in the stratosphere to remove pollutants, and the pollutants could give rise to changes in climate. 30. Melting of polar ice caps, rise of oceans; change of agriculture; etc. ANSWERS TO VISUAL CONNECTION (a) rain

(b) snow

(d) hail

(e) dew

(f) frost

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. In general, fair weather is associated with high pressure, since the pressure reduces the formation of clouds and hence precipitation. Bad weather is associated with low pressure, which allows cloud formation and precipitation. 2. Alternative energy sources that do not produce CO2, such as nuclear generating stations and solar energy. 3. Claudette.

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4. Lightning, about five times. ANSWERS TO EXERCISES 1. See Figure 20.4. 2. (a) cT (b) mA (c) cA (d) mT (e) cP and cA. 3. dc = vct = (35 km/h)(24 h) = 840 km (521 mi), dw = vwt = (20 km/h)(24 h) = 480 km (298 mi) 4. Personal. t = d/v, where average v from Exercise 3 may be used. 5. d = vt = (1/3 km/s)(10 s) = 3.3 km

(1/5 mi/s)(10 s) = 2.0 mi

6. d = vt = (1/5 mi/s)(6 s) = 1.2 mi

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Chapter 21 STRUCTURAL GEOLOGY AND PLATE TECTONICS Our present conception of the Earth’s structural geology began when geologists directed their studies to the oceans in the 1950s. The measurement of ocean depths and the charting of the topography of the ocean floor revealed a system of mid-oceanic ridges that extended through various oceans. In 1952, Harry H. Hess of Princeton University introduced the concept of convection currents and production of the ridges by rising magma from the Earth’s deep interior, which in turn cause the continents to move. This new theory was called seafloor spreading. Five years later, Jason Morgan of Princeton presented to the American Geological Union the concept that the surface of the Earth consists of rigid plates that move with respect to each other. Verification of Alfred Wegener’s early theory (1915) of continental drift was finally concluded. Our discussion of structural geology begins with the overview of the Earth’s interior structure, then and introduction to the concepts of continental drift and seafloor spreading. This is followed by plate tectonics as the primary mover and shaker of the Earth’s outer shell, the lithosphere. Crustal motion on other planets and moons is discussed in a highlight. The large majority of volcanoes and earthquakes are associated with the movement of lithospheric plates; therefore, they are our next topic for discussion. Section 21.5 presents the causes for earthquakes and safety measures to be taken before, during, and after an earthquake. Conceptual questions and answers emphasize earthquakes, and highlights focus on earthquake risks and the deadly byproducts of earthquakes: tsunamis. The forces that build up in the vicinity of plate boundaries result in crustal deformation and mountain building. The chapter ends with a discussion to these two topics.

DEMONSTRATIONS The concept of isostasy can be demonstrated using large and small ice cubes in a glass.

ANSWERS TO MATCHING QUESTIONS a. 1 b. 19

c. 25 d. 11

e. 12

j. 28

k. 7 l. 13 m. 8 n. 15

x. 24

y. 23

z. 21

f. 16

g. 6

o. 18 p. 26

h. 3

i. 2

q. 22 r. 9

s. 17

t. 10 u. 27

v. 14

w. 5

aa. 4 bb. 20

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ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b 2. d 3. c 4. d 5. a 6. d 7. c 8. c 9. c 10. c 11. a 12. b

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. Geology 2. surface waves 3. asthenosphere 4. outer core 5. deep-sea trenches 6. mid-ocean 7. isostasy 8. transform 9. ring of fire 10. Richter scale 11. volcanic 12. reverse

ANSWERS TO SHORT-ANSWER QUESTIONS 1. 6400 km, 4000 miles. 2. Crust, mantle, outer core, inner core. 3. The inner core is about 6000°C (10,800°F). 4. The asthenosphere is plastic and able to move, allowing plates above it to shift. 5. The mechanism behind continental drift is seafloor spreading most noticeable at the mid-ocean ridges. 6. The Mid-Atlantic Ridge is a ridge that runs along the center of the Atlantic Ocean between the continents. 7. Continuity of geologic features, biological evidence, glacial evidence. 8. Remnant magnetism gives the long-term record of the magnetism of the Earth. It can indicate the direction of the Earth’s magnetic field, show that Earth’s magnetic field undergoes reversals, and supports the mechanism of seafloor spreading. 9. A huge slab of lithosphere material. 10. The lithosphere is the solid outer shell of the Earth; it includes the crust and some of the upper mantle. The asthenosphere is the region below the lithosphere that is hot enough to be easily deformed and is capable of internal flow. 11. The concept that the Earth’s crustal material “floats” in gravitational equilibrium on a “fluid” substratum. 12. Divergent (Red Sea); convergent (Himalayan Mountains); transform (San Andreas Fault). 13. The region where one plate is deflected downward and beneath another plate. 14. The primary force that moves plates is gravity.

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15. The oceanic plate is subducted. Oceanic crust has a higher density (3.0 g/cm3) than does continental crust (2.7 g/cm3). 16. Volcanoes are produced primarily at plate boundaries because of the stresses caused as plates shift. 17. Volcanoes on Earth are generally found at the boundaries of tectonic plates. Specifically they are found near convergent plate boundaries. 18. Three examples of mountains formed by volcanic activity are: 1) Japan islands, 2) the Andes Mountains, and 3) Cascade Mountains. 19. Earthquakes generally occur at plate boundaries. 20. The Pacific plate is moving northward relative to the North American plate. 21. The focus of an earthquake is the point or region of the initial energy release or slippage. The epicenter is the location on Earth’s surface directly above the focus. 22. Deadly tsunami waves can travel thousands of miles. 23. (a) The Richter scale gives an absolute measure of the energy released by calculating the energy of seismic waves at a standard distance. (b) See Table 21.1 for the effects of earthquake damage. 24. Quake magnitude, location of the focus and epicenter, and environment of the region around the epicenter. 25. An anticline is a rock fold with downward sloping on both sides of a common crest. A syncline is a rock fold with upward sloping on both sides of a common trough. 26. (a) Normal fault: occurs as the result of expansive forces that cause its overlying side to move downward relative to the side beneath it. (b) Reverse fault: occurs as the result of compressional forces that cause the overlying side of the fault to move upward relative to the side beneath it. (c) Transform, or strike-slip fault: occurs when the stresses are parallel to the fault boundary such that the fault slip is horizontal. 27. A fold mountain is characterized by folded rock strata; they also show external evidence of faulting and internal evidence of high temperature and pressure changes. They have thick sedimentary strata indicating that they were once at the bottom of an ocean basin. 28. (a) Fault-block. (b) Fold. (c) Volcanic. (d) Fold.

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ANSWERS TO VISUAL CONNECTION (a) mantle, rock; (b) crust, rock; (c) outer core, iron and nickel; (d) inner core, iron

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Both use the process of investigation to obtain and identify clues to discover the unknown. The geologist investigates for clues to understand the Earth’s past and present. The detective investigates for clues to solve a crime. 2. The interior of Earth is made up of heavier elements. We know this because we can determine the overall average density of the Earth by knowing its mass and volume. Since the densities of the rocky crust, the liquid water, and the ice on the surface of the Earth are known, the interior must be denser to agree with the calculated average density of the Earth. The mass of the Earth = 5.94 x 1024 kg = 5.94 x 1027 g The volume of the Earth V = 4πr3/3 = 1.1 x 1021 m3 = 1.1 x 1027 cm3 Thus, the density ρ = m/V = 5.46 g/cm3. This is 5.46 times denser than water. 3. Tectonic plate motions are accurately measured using space-based methods. That is, scientists use satellites to carefully measure the motions of each plate relative to a frame of reference. The global positioning system (GPS) measures the movement of ground stations and has allowed scientists to make vast improvements in the accuracy of the plate motions. 4. Oil and gas deposits are likely found at or near subduction zones. 5. Individual answers will differ. 6. The two continents on Earth that have the fewest earthquakes and volcanoes are Australia and Antarctica. On which of these two continents would you live? 7. Tectonic forces caused by gravity, pressure and friction forces between rocks and plates are the main forces involved in mountain building. Other factors include the density, composition, and temperature of the rock: certain rocks under high stress will deform by faulting (fracturing), but other rocks will buckle and fold (folding) under high stresses.

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Chapter 22 MINERALS, ROCKS, AND VOLCANOES The fundamental principle underlying most geology is simply that the processes occurring at present on the Earth have occurred throughout geologic time. Thus ancient rocks can be interpreted with respect to present processes. The composition of rocks is expressed in terms of minerals, which are the building blocks of rocks. Our study of geology begins with the study of minerals. Mineral classification is based on physical and chemical properties. The physical properties distinguish different forms of minerals with the same composition. In addition, physical properties provide a convenient means of mineral identification. Although the number of common minerals is small—fewer than 20 make up over 95% of the rocks in the Earth’s crust—the student may find the number (and names) overwhelming. As a result, only a few minerals are examined in detail. Among these is silicon dioxide (SiO2), which occurs abundantly as quartz, sand, and many other minerals. Following the basic information concerning minerals and igneous rocks, we consider the igneous activity that produces the majority of rocks that form the Earth’s crust. The chapter concludes with sections on sedimentary rocks and metamorphic rocks. The chapter Highlight discusses how volcanoes can produce gold in rocks. The conceptual questions and answers focus on some interesting applications and the broad context of geologic processes.

DEMONSTRATIONS The best demonstrations for this chapter are representative mineral and rock collections that students can examine and analyze. Also, use rock charts that illustrate types, characteristics, identifying features, and the interrelationships of various types of rock. A chart showing the rock cycle should be a continuous exhibit in the classroom and laboratory during the teaching of geology. The chart on the chemistry of rocks is a good demonstration exhibit.

ANSWERS TO MATCHING QUESTIONS a. 24 b. 15 n. 20

c. 16

d. 21

e. 17

f. 9

g. 2

h. 18

o. 22 p. 1

q. 14

r. 5

s. 23

t. 3

u. 8

i. 19 j. 6 v. 11

w. 13

k. 10

l. 12

x. 4

y. 25

m. 7

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ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d 2. b 3. a 4. c 5. b 6. b 7. b 8. a 9. b 10. b 11. c 12. a 13. 14. c 15. d 16. d 17. a 18. a 19. d 20. b

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. silicon 2. feldspars 3. luster 4. gem 5. igneous 6. uniformitarianism 7. pressure 8. basalt 9. magma 10. viscosity 11. cinder cones 12. concordant 13. hot spot 14. pluton 15. gas 16. coal 17. stalactites 18. slate 19. Foliation 20. hydrothermal

ANSWERS TO SHORT-ANSWER QUESTIONS 1. A mineral is a naturally occurring, crystalline, inorganic substance (element or compound) that possesses a fairly definite chemical composition and a distinctive set of physical properties. The study of minerals is called mineralogy. 2. The silicon-oxygen tetrahedron is a covalently bonded structure composed of one silicon atom surrounded by four oxygen atoms. The structure is the basic building block of silicate materials. See Figure 21.2a for a sketch of its structure. 3. The oxygen-to-silicon ratio in silica is 2-to-1. In other silicate rocks this ratio is greater than 2 and it can vary significantly. 4. Nonsilicate materials include the pure elements like gold and silver, gemstones like diamonds and sapphires, and ores like iron, copper, and nickel. 5. The Mohs scale has limits of 1 (soft) to 10 (hard). 6. Diamonds are initially cut by strikes from steel blades along their weakest structural planes, the tetrahedral planes. For fine cuts and polishing, diamond saws and polishing wheels coated with diamond dust are used. 7. Luster is the appearance of a mineral’s surfaces in reflected light. Streak refers to the color of the powder of a mineral. 8. Mica has one perfect cleavage plane, forming sheets. 9. A rock is defined as a solid, cohesive, natural aggregate of one or more minerals. The study of rocks is called petrology.

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10. Igneous, formed at or below the Earth’s surface from molten magma or lava. Sedimentary, formed at the Earth’s surface from cemented fragments of preexisting rocks. Metamorphic, formed below Earth’s surface when the structure and mineral content of a rock are changed while the rock remains a solid. 11. Igneous rock can become metamorphic rock by heat and pressure (metamorphism), and this metamorphic rock can return to igneous rock by melting, cooling, and solidification. 12. Sedimentary rocks must undergo heat and pressure to form metamorphic rock first, then they can be melted and cooled to become igneous rock. 13. The molten material is known as magma as long as it is beneath the Earth’s surface but becomes lava if it flows on the surface. 14. Intrusive rock is igneous rock that formed below Earth’s surface and has a coarse-grained texture. Extrusive rock is igneous rock formed outside Earth’s crust and has a fine-grained texture. 15. Igneous rocks have three textures, coarse, fine, and glassy. Examples of coarse texture igneous rocks are granite, fine texture is basalt, and glassy texture is obsidian. 16. Gabbro has a larger grain size, darker color, and lower silica content than does rhyolite. 17. Plutons are solidified magma that forms below the surface while lava is the molten rock that flows on the surface. When lava cools it will form solidified igneous rock. Plutons are different shapes and sizes and depend on the types of rock they penetrate. 18. The three products of a volcano are gas, lava, and pyroclastics (solids). 19. Chemical composition and temperature. 20. The Hawaiian Islands formed from the Pacific plate moving over a hot spot. 21. Volcanoes can erupt either peacefully or explosively. The type of eruption largely depends on the viscosity and temperature of the magma inside it. Low viscosity, high temperature magma flows easily and peacefully. High viscosity, low temperature magma, however, can get stuck in vents and chambers which allow pressures to build up and potentially explode. 22. A shield volcano has a gently sloping, low profile. A stratovolcano has a steeply sloping, layered composite cone. A cinder cone volcano is steeply sloping and composed primarily of pyroclastics (tephra). 23. Caldera is a term applied to a roughly circular, steep-walled depression near the summit of a volcano. It may be up to several kilometers in diameter. Crater Lake is in a caldera.

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24. Coal is the organic chemical sedimentary rock. While minerals are not transported in solution to make the rock, coal is classified as sedimentary because it is the lithified remains of plant matter. 25. Stalactites are formed from calcium carbonate precipitating from dripping water from the ceiling of caves. Stalagmites are built up from the floor from the dripping water. You can remember stalactites extend down from the ceiling (c for ceiling) and stalagmites are built up from the ground (g for ground). 26. Physical characteristics of sedimentary rock include color, rounding, sorting, bedding, fossil content, ripple marks, mud cracks, footprints, and raindrop prints. 27. Sedimentary rocks often contain fossils. 28. Temperature and Pressure both increase with depth into the Earth. Temperature increases at a rate of about 30 C°/km, while pressure in the crust increases about 300 atmospheres per kilometer. At these high temperatures and pressures inside the Earth metamorphic rock is formed. 29. Contact metamorphism is change brought about primarily by heat, with very little pressure involved. Shear metamorphism changes rock more by pressure than temperature. Hydrothermal metamorphism changes rock by the circulation of chemically active solutions. 30. Foliation is the ability of a metamorphic rock to split along a smooth plane. 31. Verde Antique is a valuable metamorphic rock because it has a beautiful, dark green color with whitish accents. It is also durable, resists staining, and keeps its shine.

ANSWERS TO VISUAL CONNECTION (a) heat/pressure, (b) melting, (c) cooling, (d) weathering/erosion, (e) compaction/cementation

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. It is uncommon to find fossils in igneous rock because that type of rock was heated and melted. Therefore any fossils would have been destroyed in the making of the rock. Most fossils are found in sedimentary rock. 2. A Mauna Loa eruption involves basaltic magma, which has a high temperature and a low silica content. This provides a low viscosity, and thus magma and lava flow fluidly. The eruption will be peaceful. Mount Saint Helens erupts explosively because of high silica content and cooler

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temperature. Instead of flowing out the vent, a resistant plug is formed. The pressure builds up under the plug, and when the plug can no longer hold, the volcano erupts violently. 3. In order to identify the materials, you can see which mineral scratches the other and rank them: Calcite crystal (3.0), window glass (5-6), quartz, (7.0), zircon (7.5), and diamond (10). 4. Hematite. 5. There are many good websites that give tests and characteristics of meteorites. The best method to identify meteorites is to have eye witnesses verify their fall through the atmosphere. If this is not possible, then several tests can be done to prove it is of extraterrestrial origin. Meteorites differ from Earth rock in that they have a fusion crust, they often contain smooth depressions in their surfaces called regmaglypts, are usually magnetic, contain some high density metals like iron and nickel, and contain small, round, mineral features called chondrules. Many times an expert is needed to discern whether an unknown rock is truly a meteorite. 6. Mount Fuji is located in Japan and it is classified as a stratovolcano. Mount Kilimanjaro, located in Africa (specifically Tanzania), and Mount Vesuvius, located in Italy, are both also classified as stratovolcanoes. 7. The igneous formation shown is a dike.

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Chapter 23 SURFACE PROCESSES Some internal processes, such as volcanism and mountain building, build up the Earth’s surface features. Many surface processes work in the opposite direction, wearing away and leveling surface features. The weathering and erosion of surface materials are important geologic processes, as is mass wasting, which is the downward movement of weathered materials under the influence of gravity. Evidence of these processes is commonplace. However, some processes take place very slowly and are often overlooked. Erosion is carried out by the so-called agents of erosion—running water, ice, wind, and waves—which are the physical phenomena that supply the energy for the removal and transportation of rock debris. Two of these agents, running water (streams and rivers) and ice (glaciers), will be examined in some detail. Not only is water an important factor in geologic processes, it is also necessary to sustain life and is a major environmental concern. In this chapter, the Earth’s water supply is studied. This supply is often thought to be inexhaustible, because about 70% of the Earth’s surface is covered by water. However, only about 2% of this is fresh water, and most of this 2% is locked in glacial ice sheets. The Earth’s water supply is a reusable resource that is constantly being redistributed. In general, atmospheric processes move moisture from large oceanic reservoirs. The water eventually flows back to the seas, eroding as it goes. Groundwater is an important aspect of this gigantic hydrologic cycle, particularly for domestic water supplies. The geologic features of wells and springs, as well as the quality of water with respect to dissolved minerals, are considered. Highlights on Earth’s largest crystals and Earth’s highest tides show the power of water and time. Because oceans cover about 70% of the Earth’s surface, they are important factors in surface processes. The movement of seawater affects regional climates and is an agent in the erosion of coastlines. The final topic of the chapter deals with the topology of the seafloor. Recent investigations have greatly expanded our knowledge of the surface features of this major unseen portion of the Earth’s surface.

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DEMONSTRATIONS Geology models show geologic landforms in three dimensions. Stream tables use flowing water and rainfall to show the effects of water on land-formation processes such as erosion and sedimentation. Topographic relief maps can be used to show landform features, and contour models illustrate the fundamentals of contour mapping. These models are also very useful in the laboratory when the student is asked to draw contour maps.

ANSWERS TO MATCHING QUESTIONS a. 17 b. 14

c. 20

d. 6

e. 18

f. 9

g. 23

h. 13

n. 8

p. 10

q. 16

r. 7 s. 15

t. 19

u. 2

o. 1

i. 11

j. 3

k. 21

v. 22

w. 12

x. 5

l. 4

m. 24

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. a 2. c 3. c 4. c 5. d 6. b 7. b 8. c 9. d 10. b 11. b 12. a

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. mechanical 2. permafrost 3. dissolution 4. stream 5. bed 6. continental glaciers 7. mudflows 8. moraines 9. Subsidence 10. aquifer 11. guyots 12. continental slopes

ANSWERS TO SHORT-ANSWER QUESTIONS 1. Mechanical weathering involves the physical disintegration of rock (frost wedging, crystal growth, root penetration, and others), and chemical weathering involves chemical changes in rock composition (dissolution, oxidation, and hydrolysis). 2. Chemical weathering is most prevalent in hot, moist climates 3. Burrowing animals loosen soil and bring it to the surface. Plant root systems fracture rocks. 4. Permafrost is a layer of subsurface soil that remains frozen permanently. 5. Limestone. 6. A depression resulting from the collapse of a cavern that has been formed by chemical weathering. 7. The natural influence of erosion is gravity (mass wasting). The agents of erosion are wind, streams, waves, and glaciers. 8. Dissolved, suspended, and bed loads. See Section 23.2.

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9. A loop-like bend in a river channel is a meander caused by gravity and the rotating Earth. Oxbow lakes. 10. Floodplains contain very rich soil for agriculture, however they flood regularly. 11. Continental glacial ice sheets cover large areas and flow outward. Valley glaciers form in valleys and flow down the valleys. Cirque glaciers are small glaciers that form in depressions along mountains. 12. Glacial drift is material deposited by glacial ice. 13. Earth’s largest deserts, surprisingly, are found in Antarctica and the Arctic, near the poles. 14. (a) Rockslide is a type of landslide occurring in mountain areas when large quantities of rock break off and move rapidly down steep slopes. Fast. (b) Creep is a type of slow mass wasting involving the slow, particle-by-particle movement of weathered debris down a slope, which takes place year after year. Slow. (c) Slump is a type of landslide involving the downslope movement of an unbroken block of overburden that leaves a curved depression on the slope. The overburden does not travel very fast or very far. Slow. 15. Evaporation. Precipitation. 16. Permeability is a measure of a material’s capacity to carry fluid. Porosity is the percentage volume of unoccupied space in the total volume of a substance. 17. Water wells should be drilled into the zone of saturation. 18. A body of permeable rock through which groundwater moves. Aquifers exist under more than half of the conterminous United States. 19. Groudwater extraction can lead to depletion, subsidence, saltwater and other contamination. 20. There are many ways in which groundwater can become contaminated. Three ways are: 1) saltwater contamination of wells caused by freshwater extraction, 2) chemical contamination from industries and homes, like phosphates, pesticides, and organic compounds, and 3) acid rain from air pollution. 21. Figure 23.24 shows three features of coastal deposition. Pocket beaches form in the lowenergy wave environment between headlands. Barrier islands extend more or less parallel to the mainland. Between a barrier island and the mainland is a protected body of water called a lagoon. Spits are narrow, curved projections of beach that extend into the sea, elongating the shoreline.

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22. A cork in the ocean bobs up-and-down as the waves pass beneath it. However, when the wave gets close to the shore, the circular wave motion is disrupted by the ocean floor and the paths are elliptical in shape. This causes the waves to break and form surf. If a cork is close enough to the shore then it will be thrown up onto the shore in the surf. 23. The tides are the periodic rise and fall of the water level along the shores of bodies of water. Both the gravitational influence of the Moon and the Sun contribute to the tides, but the Moon has more influence. 24. Continental shelves are the relatively shallow submerged borders of the continents. Continental slopes are the true edges of the continents, where the continental land masses slope downward to the floors of the ocean basins.

ANSWERS TO VISUAL CONNECTION (a) evaporation, (b) condensation, (c) evaportranspiration, (d) precipitation, (e) runoff

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Yes. The Moon has mass, and gravity can produce downslope movement of material. 2. The presence of roads and buildings do not allow rainwater to soak into the ground where it falls. The roads move the water quickly to low lying places where the water can quickly rise to flood these regions. Other subsurface regions are robbed of their water and the groundwater can become depleted. Ground water is also affected because it can easily become contaminated by human, agricultural and industrial wastes. 3. Weathering caused the rapid degradation of Cleopatra’s needle. For thousands of years it stood in the dry desert of Egypt. In only 100 years in New York City it weathered greatly from the rain, humidity, freezing, and acid rain breakdown of the rock. 4. Earth’s fresh water supply will not change significantly due to the melting of the glaciers, because most of the melting occurs in the salt water ocean. The overall liquid water supply will increase and ocean levels will rise significantly. 5. There are many way in which to protect your drinking water. Specifically one can reduce consumption and thereby reducing the freshwater extracted from the ground. Other ways include reducing or removing pesticides and fertilizers used on lawns. Reducing the use of organic

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compounds and phosphates can help keep groundwater clean. Finally, avoid improper dumping of garbage; recycle as much as possible. 6. Suspended load. Dissolved load. Bed load.

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Chapter 24 GEOLOGIC TIME This chapter ends by focusing on a topic of interest to most students—geologic time. The chapter begins with a discussion of what fossils are, how they are formed (in a Highlight), and the names of some common ones. The principles used to establish relative geologic time are then treated, and the eon, era, and period time units are explained. The next topic is radiometric dating and how it is used to establish the age of the Earth and absolute geologic time. The last section of the chapter tells how relative and absolute geologic times are combined into the geologic time scale. Another chapter Highlight discusses the evidence for an asteroid strike causing the demise of the dinosaurs. Few students come into a physical science class having anything but the haziest notions of geologic time, so we recommend that the entire chapter be covered. Two conceptual questions and answers are given to help understand fossilization and the geologic time scale.

DEMONSTRATIONS Fossil sets can be used to demonstrate the major types of fossils, their identification, and the geologic time periods they represent. Fossils actually found in nearby localities are always of interest. Students can be asked to bring in any fossils they may have found. Posters, slides, sets, kits, models, and paleoplaques of fossils and Earth history are readily available from scientific suppliers. Figure 24.20 is an excellent demonstration of the scale of geologic time. Your students can try making another one by correlating a distance scale with a time scale (1 mi or 1 km equals one million years of Earth’s history).

ANSWERS TO MATCHING QUESTIONS a. 15 b. 24

c. 22

d. 21

e. 8 f. 9 g. 2

h. 17

i. 26

j. 19

k. 6

n. 5

p. 10

q. 13

r. 16

u. 18

v. 3

w. 11

x. 14

o. 4

s. 1

t. 20

l. 23 y. 25

m. 7 z. 12

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d 2. c 3. c 4. b 5. a 6. b 7. d 8. b 9. a 10. d 11. d 12. a 13. b 14. c 15. d 16. d 17. c 18. a 19. b 20.c

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ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. mechanical 2. replacement fossil 3. algae 4. Cenozoic 5. unconformity 6. cross-cutting relationships 7. relative time 8. Correlation 9. eons 10. tribolite 11. superposition 12. carbon-14 13. more 14. primordial 15. 4.56 16. younger 17. Rodinia 18. K-T Event 19. explosion 20. 100,00

ANSWERS TO SHORT-ANSWER QUESTIONS 1. Geologic time is the time span since the formation of planet Earth. 2. Worms do not have a hard skeleton and thus decompose quickly before a fossil can form. 3. The fossil record shows that as time passed, larger and more complex life forms developed. 4. Examples of replaced remains include petrified wood and carbonization of plants. 5. Petrified wood. 6. Coal is formed from the carbonization of plants. This process involves the decomposition of plant remains by bacteria under anaerobic conditions. 7. When an embedded shell or bone is dissolved completely out of a rock, it leaves a hollow depression called a mold. If new mineral material fills the mold, it forms a cast of the original shell or bone. 8. Fossils of blue-green algae, or cyanobacteria, date back to about 3500 mya (million years ago). 9. Layers of rock that contain certain microfossils, such as some species of Foraminifera, indicate the presence of nearby oil deposits. 10. Three examples of trace fossils are tracks, borings, and burrows. 11. In a sequence of undisturbed sedimentary rocks, lava, or ash, each layer is younger than the layer beneath it. Example: If lava A covers rock layer B, A is younger than B. If the rock layers are disturbed by folds or mountain building then the principle may be violated. 12. No, non-horizontal rock layers were originally horizontal and were altered later by other forces.

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13. Faults are used in relative aging because a fault must be younger than any of the rocks it has affected. 14. An unconformity is a gap, or break, in the rock record at a given locality due to nondeposition or erosion. 15. Correlation is the process of matching up rock layers in different localities by use of index fossils or other means. 16. Index fossils aid correlation because they are found in a limited time segment of Earth’s history. 17. Typical of a particular limited time segment, widespread, numerous, easily identified. 18. Events in your daily life can be an analogy to relative geologic time. In relative time, you know the order of events, but you do not know the actual time. Examples from daily life include waking, eating breakfast, going to a meeting at work, picking up children from school, and watching the news on TV. 19. Paleozoic, Mesozoic, Cenozoic. 20. Seven, Permian. 21. Triassic, Jurassic, Cretaceous. 22. Tertiary, Quaternary. 23. Carbonaceous period. 24. Precambrian time is the huge span of time before the Cambrian period of the Paleozoic era, that is, the time before the Phanerozoic eon. 25. The oldest rocks are dated by rubidium-strontium radiometric dating. 26. Daughter products that are gaseous can leak out of the rocks being dated. 27. Carbon-12. 28. No addition or subtraction of the parent or daughter has occurred over the lifetime of the rock other than that caused by radioactive decay. The age of the rock does not differ too much from the half-life of the parent radionuclide. None of the daughter element was present in the rock when it was formed, or if it was, it is possible to tell how much. 29. Lead that comes from radioactive decay is called radiogenic lead, whereas lead that does not is called primordial lead.

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30. Carbon-14 dating is used to date objects within about 7000 years old. These include the ages of organic material from bones, charcoal from ancient fires, and the wood beams in the pyramids. 31. Meteorites from the asteroid belt date back to 4600 mya., the oldest Moon rocks date back 4550 mya., and the oldest known Earth rocks date back 4000 mya. All these parts of the solar system are thought to have formed about the same time. 32. The oldest known Earth rocks date back 4000 mya.; the oldest Moon rocks date back 4550 mya. 33. By dating igneous rocks that intrude or cover sedimentary layers and then using the law of superposition. 34. The Hadean eon starts at Earth’s formation; the Archean eon starts at earliest known Earth rocks; the Proterozoic eon starts at the formation of the North American continental core. 35. Cenozoic (65 million years). 36. Cenozoic era, Quaternary period, Holocene epoch. 37. Asteroids, supernova explosions, greenhouse warming, or ice-age cooling. 38. The end of the Ice Ages (last retreat of glaciers from North America and Europe). 39. Earliest fossils of the genus Homo. 40. The Cambrian explosion. 41. The K-T event. 42. Unusually high concentrations of iridium, shocked quartz, and soot in layers of rock from 65 mya. indicate an asteroid impact. A large impact crater is found at Chicxulub that dates from 65 mya. (Any two pieces of evidence are a sufficient answer.) 43. The K-T event has evidence that points to an astronomical origin. Evidence includes the rich amount of iridium found in the Earth’s soil at the 65 million year old boundary. Iridium is prevalent in meteoritic rock but not generally found in Earth rock. Also shocked quartz rock, spherules, and soot were found at the K-T boundary. This evidence points to a catastrophic event at 65 million years old followed by worldwide fires, dust, ash, and debris. Core drilling also provided evidence of pieces of the carbon-iron-rich meteorite itself. 44. Coal forming forests were alive in the Paleozoic era and Pennsylvanian period. 45. Dinosaurs lived about 14 days on the geologic time calendar; this corresponds to about 165 million years of time.

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ANSWERS TO VISUAL CONNECTION (a) Parent 30, Daughter10; (b) Parent 20, Daughter 20; (c) Parent 10, Daughter 30; (d) Parent 5, Daughter 35; (e) Parent 2.5, Daughter 37.5

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. The radiometric dating technique is useful because it can be applied to a wide array of rocks over a large range of time. Many different isotopes can be used to cross-check, calibrate, and validate results, thus reducing errors. One weakness of the method is that the daughter products must be of high enough quantity to make an accurate age estimate. Certain ages of rock require radioactive elements with similar half-lives. Another weakness is that it is sometimes difficult to know whether a rock has any daughter element that was present when it formed. 2. All three objects lived in the Paleozoic era which spans from about 550-250 million years old. Crinoids come from the Mississippian period; they are about 340 million years old. Since all three were found in the same rock, a good estimate of the age is 340 million years old. 3. The last of the dinosaurs died out 65 mya., whereas the oldest Homo fossils known only date to about 2 mya. So a huge 63-million-year gap seems to exist between the last dinosaurs and the earliest humans. 4. The amber itself is prehistoric petrified tree resin, so it is a fossil whether or not an insect is embedded in it. 5. Dinosaurs lived from about 245 to 65 mya., but carbon dating can only date materials younger than about 75,000 y. Perhaps carbon dating was confused with another type of radiometric dating.

ANSWERS TO EXERCISES 1. (a) Mississippian, Pennsylvanian, Permian. (b) Silurian.(c) Platycrinites and Lingula. (d) Elrathia. 2. (a) Ordovician and Silurian. (b) Devonian. (c) Phacops, Zygospira, and Lingula. (d) Neither lived during that period. (e) Elrathia. 3. (a) A is the younger; superposition. (b) E is the younger; cross-cutting relationships. 4. (a) D is the younger; cross-cutting relationships. (b) E is the younger; cross-cutting relationships.

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5. About 11,460 years old. 6. 3.9 billion years. 7. Older than 210 my but younger than 245 my. 8. The rock must be between 410 my and 350 my old, the time period during which both fossils lived.

Part 2

AN INTRODUCTION TO

Physical Science Instructor’s Guide to accompany Laboratory Guide

Shipman, James Wilson, Jerry Higgins, Charles

Prepared by

Shipman, James Baker, Clyde

CONTENTS Experiments: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

Graphs Measurement The Simple Pendulum Uniform and Accelerated Motion Determining g, the Acceleration of Gravity Newton’s Second Law Hooke’s Law for a Vibrating Spring Centripetal Acceleration and Force Laws of Equilibrium Principle of Work Using an Inclined Plane and Pulleys Waves Interference of Light Waves Plane Mirrors and Index of Refraction of Light Mirrors, Lenses, and Prisms The Refracting Telescope Color Static Electricity Magnetism and Electromagnetism Ohm’s Law Electric Circuits Electromagnetic Waves Temperature Specific Heat Heat of Fusion Heat of Vaporization of Water Radiation Spectroscopy Density of Liquids and Solids Oxygen Percentage of Oxygen in Potassium Chlorate Percentage of Oxygen and Nitrogen in the Air An Exothermic Chemical Reaction Avogadro’s Number Molecular Structure Solutions and Solubility Pressure-Volume Relationship of Gases Chemical Qualitative Analysis Chemical Quantitative Analysis (Volumetric) Kepler’s Law Stars and Their Apparent Motions Locating Stars in the Night Sky Motions and Phases of the Moon

1 7 11 15 19 23 27 31 33 37 41 43 45 49 53 57 63 67 71 75 85 89 91 95 99 103 105 109 111 113 115 117 119 123 125 127 131 133 135 139 143 149 iii

43 44 45 46 47 48 49 50 51 52 53 54 55

Observing the Phases of the Moon Hubble’s Law Measuring the Radius of the Observable Universe Air Pressure Humidity Weather Maps (Part 1) Weather Maps (Part 2) Topographic Maps Minerals Rocks Rock-Forming Minerals Igneous Rocks and Crystallization Sedimentary Rocks

153 159 163 171 173 175 177 179 181 183 185 187 189

Experiment 1

Graphs INTRODUCTION The objective of this experiment is to introduce students to graphing and how graphs are used to illustrate relationships in science. Some students will most likely be acquainted with graphs and how they are plotted, but for others, this will be the first experience with graphs. Thus a few remarks are necessary. First, review the major points in the introduction. Second, go through the guidelines for plotting a good graph. Some students will need special help in choosing proper scales for the x and y axes. Many students have trouble compiling a complete title for the graph. Stress guideline 4. Even though full-page graphs are requested, a few students will still use only about one-fourth of the graph paper to plot the data. Third, point out that ordered pairs of numbers obtained from an equation will produce a graph where the plotted points will fall exactly on the curve connecting these points. This will not be the case when actual data from a physical or chemical experiment are used as the basis for the plotting. Remind students that mathematics is the language of science, and it must be learned and used in laboratory experiments. The concepts of direct proportion, inverse proportion, functions, ratio, and slope can become more familiar to the student through laboratory experience. Stress the construction and labeling of good graphs. Remind students to compare their graphs with the example in the Laboratory Guide. Do they have the axes properly labeled with correct units? Have they given the graph a proper name? Have they shown the calculation for the slope and recorded the units, if the slope has units? Have they included their name, date, and class in the lowerright corner of the graph? Graphs may become separated from the experiment during the grading process. Have they incorrectly used the upper and right margins of the graph for calculations? These margins should be free of marks of any kind. If these details are not given proper attention in this first experiment, many students will not produce good graphs in future experiments. Samples of the three graphs required for this experiment are included in this manual and may be copied and posted on the laboratory bulletin board for students’ reference.

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

Explain how you determined the magnitude of the divisions for both the x and y axes in order to plot a graph that utilizes the major portion of the graph paper. Ans: The answer to this question will vary. The following is one method for choosing convenient spacing. First, examine the data and determine the maximum values to be plotted on each axis. Next, determine how many divisions are needed so that the maximum value for each variable will fit on the graph. There are 10 divisions per inch on each axis, so choose spacing that will make each division an even number. This will make interpolation easier.

Distinguish between dependent and independent variables. Ans: The independent variable produces a change in the dependent variable. The dependent variable is, therefore, referred to as a function of the independent variable.

Name the dependent and independent variables in Procedure 2. Ans: The area is the dependent variable. The radius is the independent variable.

What must be included in the title used on a graph? Ans: The name of the graph that refers to the concepts plotted on the x and y axes plus the subject to which they refer must be included.

How are the units for the slope of a graph determined? Ans: The units for the slope are taken from the quantities plotted on the x and y axes. Example: slope = rise/run so on the graph for Procedure 1 the units will be distance/time or m/s.

Examine your graph for Procedure 1 and determine if the relationship between the circumference and the diameter of a circle is a linear relationship? Justify your answer. Ans: Yes, C = π d = 2π r , which is the equation for a straight line.

When the value for x increases faster than the values for y, which way (toward the y axis or away from the y axis) will the graph curve? Ans: The graph will curve away from the y axis.

State in your own words the relationship between the area of a circle and the radius2 for the same circle. Use the graph plotted in Procedure 3 to help you formulate this answer. Ans: The area of a circle is directly proportional to the radius squared, as shown on the graph prepared in Procedure 3.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

How does the slope of the graph in Procedure 1 compare with the average value of C d ? Ans: They are the same.

What shape of graph should be obtained in Procedure 2? Ans: A parabola

What shape of graph should be obtained in Procedure 3? Ans: A straight line

Does the slope of the graph in Procedure 1 have units? Why or why not? Ans: No. The units of the x and y axis are the same. They cancel.

Which variable is increasing the fastest on the graph prepared in Procedure 2? Ans: The area increases in magnitude fastest.

State the physical meaning that the origin (0, 0) represents on the graph drawn in Procedure 2. Ans: These are not measured points, but when the radius is zero, the area must also be zero.

Sample Graph for Procedure 1

Sample Graph for Procedure 2 160

140

120

100

Radius (cm)

Experiment 2

Measurement INTRODUCTION This experiment introduces students to the fundamental units of measurement (length, mass, and time) and the techniques for taking measurements, plus an introduction to the concepts of least count, significant figures, accuracy, precision, percent difference, and percent error. The experiment also provides essential information for future laboratory experiments. Each student should be encouraged to take an active, hands-on role in the measurement of all items provided in the experiment. Some students have trouble in determining the least count of a measuring instrument as asked for in Procedure 1a. To reduce the number of students’ questions, the least count of a typical meter stick has been determined as an example. The instructor should introduce the entire class to the use and care of the triple-beam balance and briefly explain how the simple pendulum is to be supported and how its length should be measured. Students will be asked to calculate percent error and percent difference in future experiments. The necessary equations are given in Appendix III of the Laboratory Guide.

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

Define least count, and then determine the least count of your (or someone else’s) wristwatch. Ans: The least count is the smallest division marked on any measuring instrument. One second is usually the least count of a wristwatch with a second hand.

Define measurement, and give an example. Ans: Measurement is the comparison of an unknown quantity to a standard. Example: The long dimension of a page of the Laboratory Guide is 27.7 cm, that us 27.7 times as long as a standard centimeter.

Differentiate between accuracy and precision. Ans: Accuracy refers to how well the experimental value agrees with the true or exact value. Precision refers to the degree of reproducibility of a measurement, that is, to the maximum possible variations in of the actual measurements taken. Both may be expressed as a plus/minus correction value or as a percentage.

Why should the experimenter avoid using the end of a meter stick when making a measurement? Ans: The end of the meter stick is usually damaged and thus becomes unreliable as a reference point.

Calculate the length in meters of a simple pendulum that has a period of 1.00 second? Ans:

T 2g 4π

= 2

(1.00 s )2 9.80 m s2 = 0.248 m 2 4 ( 3.14 )

Explain under what conditions you would use percentage error and when you would use percentage difference. Why would you sometimes use both? Ans: Percent error is defined as the ratio of the absolute difference between the experimental and the accepted values to the accepted value, expressed as a percentage. Percent difference is defined as the ratio of the absolute difference between the experimental values to an average of the experimental values, expressed as a percentage. Percent error is a measure of accuracy while percent difference relates to precision, so both can tell something valuable about how well an experiment has been done.

How can the accuracy of a measurement be increased? Ans: Take more readings to make the average better and take more care in using the measuring instruments to provide as many significant digits as possible in your data.

How can precision of a measuring instrument be increased? Ans: One way to do this is to increase the number of calibrated subdivisions (the least count) that are marked on the measuring instrument or you might choose a better type of measuring instrument all together. Example: Use a vernier caliper instead of a meter stick.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

How can the meter stick be used to measure the thickness of a single sheet of the Laboratory Guide. Ans: Measure the thickness of several pages (100 or more) at one time; then divide by the number of pages measured.

Why are several observations or trials taken for each measurement? Ans: Several trials are taken to improve the accuracy of the measurement by providing a more representative average value.

Explain the relationship between the terms significant figure and significant digit. Ans: A significant figure is the entire number that contains all known significant digits (individual integers) plus one doubtful digit.

When performing arithmetical operations using significant figures, how many significant digits can there be in a final answer that is obtained by multiplying or dividing the original recorded data? Ans: The final computed answer can contain no more significant digits than the number of significant digits in the original measurement with the least number of significant digits.

In what way does the Heisenberg uncertainty principle limit our ability to measure exactly? Ans: The Heisenberg uncertainty principle is a limit on measurement accuracy that is philosophically significant, but it is of practical importance only when dealing with measurements made on atomic or subatomic size particles. As long as the size of the object being measured is relatively large, this effect will be very, very small and can be neglected. Refer to the textbook, Section 9.5, for more details.

Experiment 3

The Simple Pendulum INTRODUCTION The simple pendulum was introduced in Experiment 2. This experiment offers a more extensive study of the pendulum, plus an additional study of nonlinear graphing. The pendulum used should be as long as possible for the first measurement. It is sometimes convenient to set up one very long-length pendulum for the entire class and then let one student group collect data and write them on the blackboard for all to copy. Also, if time is a factor, each student can collect data for one length of the pendulum and share that data with the others. The instructor must impress on students that the values for the pendulum’s length be made exactly one-half the value of the preceding length. This procedure allows students to see from the collected data that a pendulum with a length four times as long as another will have a period twice that of the shorter length pendulum. Tell students that when they start the pendulum swinging, they should not start the timing device when they let go of the pendulum bob. Allow the pendulum to swing through a few cycles before starting the timing device to allow for variations in the way the bob is released. For good results in the calculations, the period must be measured to the nearest 1/10 s. If possible, measure the period to the nearest 1/100 s. Remember that the period is squared in the pendulum equation so accurate measurements are very important. T = 2π

L g

Eq. 3.1

The first graph (period versus length) shows a nonlinear relationship as shown in the above equation (Eq. 3.1). The graph can be extrapolated through the origin, since a length of zero would have a period of zero. The second graph (period2 versus length) shows an increasing linear relationship that indicates period2 is directly proportional to the length. The slope should be approximately 0.04 s2 cm . This graph also can be extrapolated through the origin.

Some instructors put the equation (part of an infinite series) T = 2π

FG H

IJ K

L 1 θ 9 θ 1 + sin 2 + sin 4 +! g 4 2 64 2

on the chalkboard and indicate that Eq. 3.1 is only accurate for very small angles. The experimental value for the period in Procedure 3 should be slightly greater than that obtained in Procedure 1. The students may obtain a slightly greater value if they are timing to 1/100 s and are extremely careful in determining the time of 10 cycles. The following are theoretical calculations for the period and period2 for Data Tables 3.1 and 3.2 using Eq. 3.1. The theoretical data will be the same for both tables, since the period of a simple pendulum is not a function of the mass of the bob. If the student is careful in determining the time for 10 cycles, the experimental data should vary by not more than 5 percent from these values. The data for Table 3.3 may show slightly larger values for the period. See the answer to Question 5 below.

Length

DATA TABLE 3.1 (Theoretical Calculated Values using Eq. 3.1) Period

Period2

160 m

2.54 s

6.45 s2

80 m

1.79 s

3.20 s2

40 m

1.27 s

1.61 s2

20 m

0.90 s

0.81 s2

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

Why do you think you were asked to complete more cycles in Procedure 1, Step 3, if the total time was less than 10 seconds for 10 cycles? Ans: More cycles were completed to increase the accuracy of the data.

What type of graph (straight line, parabola, hyperbola) was obtained in Procedure 1, Step 5? Ans: The plot of the period of a simple pendulum versus the length will yield a parabola.

Which variable (period or length) does the graph show to be increasing the fastest? (Note: If they were increasing at the same rate, the graph would be a straight line.) Ans: The value of the length increases faster than the value of the period as shown because the graph curves toward the length axis.

What type of graph was obtained in Procedure 1, Step 6? Ans: A straight line is obtained when period squared is plotted as a function of length.

State how the period of the simple pendulum varied experimentally with the (a) length of the pendulum, (b) length of arc (displacement of the bob), and (c) mass of the bob. The period increases proportionally as the square root of the length increases. Ans: (a) The period increases slightly as the displacement angle increases. The first(b) order approximation explaining this is T = 2π

(c)

FG H

IJ K

L 1 θ 1 + sin 2 where theta (θ ) is the displacement angle. g 4 2

Theoretically, the period is independent of mass but if the length of the pendulum is not measured correctly and with great accuracy this may not be obvious from the data taken.

A simple pendulum has a length L and a period T. If the length is increased to 4L, how will be the new period be affected? Refer to data tables for the answer. Ans: The new period will be 2T, that is it will take twice as long for each swing.

Using Equation 3.1, calculate the period of your pendulum for a length of 160 cm or for the maximum length you used. How does this compare with your experimental value? Calculate the percent error. Use the calculated value of the period as the standard value. Ans: When L = 160 cm, T = 2.54 s. How well this compares to your experimental value will depend on how carefully you did your work and the percent error will show that.

Assuming the length remains the same, calculate the period of the same simple pendulum on the surface of the Moon? [Note: The acceleration (g) due to gravity on the Moon is one-sixth of that on Earth.] Ans: The period would be greater, because the value of g on the moon is less. Eq. 3.1 shows that if g = 163 cm s2 , the period on the moon would be 6.22 s.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Calculate the period of a pendulum with a length of 1.00 m when placed on the surface of the moon. The acceleration (g) due to gravity on the moon is 1/6 the value on Earth. L 100 cm = 2 × 314 = 4.92 s T = 2π . Ans: 980 cm g 6

What would be the period of your simple pendulum located in a spacecraft that is orbiting Earth? Ans: Because g is zero, the period is zero, the pendulum would not swing back and forth at all.

The period of a simple pendulum depends primarily on (a) . and (b) Ans: (a) length (b) acceleration due to gravity 13

Is it feasible to use the physical properties of a simple pendulum to designate the standard values for any of the fundamental quantities? Justify your answer. Ans: Yes, length and time could be designated with a pendulum but mass could not. The simple pendulum equation does not include mass.

Experiment 4

Uniform and Accelerated Motion INTRODUCTION The objectives of this experiment and Experiment 5 are to introduce the concepts of distance, displacement, velocity, and acceleration for straight-line motion and to use graphing techniques to analyze the relationships between these concepts. A linear air track is used to study these concepts, but should one be unavailable, a 2 in × 6 in × 10 ft long, straight wood plank with a V-groove cut the full length of the 2-in side plus a 3/4-indiameter steel ball will be satisfactory for obtaining the data. Briefly explain and give examples of the difference between distance traveled and displacement before Procedure 1 is started. Indicate that the distance traveled and the displacement are the same in this experiment since it is one dimensional motion. Students will need one or more practice trials to acquaint themselves with taking time measurements. They must obtain the data during one run of the glider or steel ball since the track or board is level during this procedure. In this experiment, Procedure 1 data are a function of how much force is applied to start the glider or ball. Procedure 2 data are a function of the adjusted air track angle. Because the data may vary, no sample data are provided for the experiment. Also, air tracks come in different lengths, and the distances to use in Data Tables 4.1 and 4.2 must be chosen by the instructor accordingly. Write the distances to be used on the blackboard if they will not be the same as those suggested in the laboratory guide.

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

Define velocity, and give an example. Ans: Velocity is the time rate of change of displacement. Ex: a car is traveling north at 60 km/h.

Define acceleration, and give an example. Ans: Acceleration is the time rate of change of velocity. Ex: a ball falls with an acceleration of 9.8 m s2 when released near the surface of the Earth.

How does the distance traveled vary with time in uniform linear motion? (Hint: Examine Eq. 4.1 in the Laboratory Guide.) Ans: The distance traveled is directly proportional to the time.

How does the distance traveled vary with time in uniform accelerated motion? (Hint: Solve Eq. 4.3 in the Laboratory Guide for d.) Ans: The distance traveled is proportional to the time squared. This is not a linear relationship so a straight line will not be obtained is d is plotted as a function of t.

In Procedure 1, what is the relationship between the velocity and the slope of the curve? (Hint: What are the units for the slope on the graph that you plotted?) Ans: The slope of this curve is exactly equal to the velocity in both magnitude and units.

In Procedure 2, what is the relationship between the acceleration and the slope of the curve? (Hint: How does Eq. 4.2 in the Laboratory Guide for acceleration compare with the equation given for slope?) Ans: The slope of this curve is exactly equal to the acceleration.

What kinematic quantity is indicated by the speedometer of an automobile? Ans: Instantaneous speed.

Give an example of an object having a velocity but no acceleration. Ans: The glider in motion on the air track in Procedure 1. Any object moving in a straight line at a constant speed.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Define instantaneous velocity. Ans: At a given instant the ratio of displacement to time is called instantaneous velocity.

What do we mean by uniform linear motion? Ans: If the ratio of displacement to time remains constant, the motion is uniform as long as the motion is also in a straight line.

A car starting from rest reaches a velocity of 80 km/h in 10 s. (a) Calculate the average velocity of the car. How many meters did the car travel in 10 s? (b) v f + v0 80 km h + 0 v= = = 40 km h Ans: (a) 2 2 40 km 10 s 40 km 1 h/ 10 s (b) d = vt = × = × × = 0110 . km = 110 m h 1 h/ 3600 s 1

Calculate the average acceleration of an automobile that attains a velocity of 60 mi/h in 10 s starting from rest. v f − v0 60 mi h 88 ft s = = = 8.8 ft s 2 a= Ans: t 10 s 10 s (Here we use the fact that 60 mi h = 88 ft s )

Is it possible for an object to have an acceleration of zero and a numerical value for velocity? Justify your answer with an example. Ans: Yes, an object moving in a straight line at constant speed will have zero acceleration.

Is it possible for an object to have a velocity of zero and a numerical value for acceleration? Justify your answer with an example. Ans: Yes, a ball thrown upward at its maximum height will have an instantaneous velocity of zero but will still have an acceleration (due to gravity) of 980 cm s2 .

Experiment 5

Determining g, the Acceleration of Gravity INTRODUCTION This experiment is a continuation of Experiment 4. The objectives are similar, and the same apparatus is used to obtain the data. Show students that the acceleration of gravity is difficult to obtain with a free-falling object because the time of drop is so short that it is difficult to measure. Hold a golf, tennis, or other small ball or object 1 m above the laboratory floor and ask the students to guess or measure the time of fall. They will give a variety of answers. Next, point out how the inclined plane slows down the motion so that more accurate measurements can be made. Most, but not all, students will have read the introduction to the experiment and understand the derivation of Eq. 5.2 in the Laboratory Guide. Therefore, the instructor may wish to derive Eq. 5.2 on the chalkboard so that all will see the reasons for the measurement of distance and time. Students should now be plotting very good graphs. If not, toughen the grading. Require students who turn in poor graphs to turn in new graphs at the next laboratory period. This makes more work for the instructor, but the student soon gets the point. The following data are typical for this experiment when the time is measured to 0.01 s. The data are for a height of the incline of 30 cm and a length of 300 cm. With this elevation, the angle θ will be approximately 6°. Distance Traveled (d), cm 50 100 150 200 250

Average Time (t), s 1.01 1.43 1.75 2.02 2.26

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

What do you think is the greatest source of error in this experiment? Justify your answer. Ans: The greatest source of error is in determining the times to travel the various distances because of the number of different students taking measurements and the human errors made by each due to their reaction times in starting and stopping the clock. 19

What kinematic concept (distance, velocity, acceleration, etc.) of the moving object does the slope of the graph represent? Ans: The slope of the graph is equal to the acceleration.

Calculate the slope of the “distance vs time squared” graph on the graph page itself. How does the value compare to the value for the acceleration found in calculation 2? Ans: The value of g will be greater than the value obtained in the experiment. The acceleration a in Eq. 5.2 in the Laboratory Guide will be greater because the value d will be greater if an initial velocity is given to the glider.

If the initial velocity of the glider or ball was not zero, how would your experimental value of g compare with the value you obtained by assuming the initial velocity was zero? For instance, suppose you accidentally gave the object a slight shove down the incline at the starting point. Would your calculated value for g be larger or smaller? Ans: Since we are assuming the initial velocity to be zero in our method of calculation, any initial velocity of the glider or ball down the track would lead to a lower value for the acceleration calculated in this way.

What would be the value of the acceleration when θ (theta) in Fig. 5.2 in the Laboratory Guide is increased to 90°? (Draw a small sketch to help you visualize this.) 980 cm s2 , the glider (or ball) would act just as if it were in freefall. Ans:

Is the acceleration due to gravity always pointing vertically downward even for an object whose velocity is vertically upward? Explain your answer. Ans: Yes. The force of gravity is always acting vertically downward, and the acceleration is always in the direction of the unbalanced force.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Why does a small error in t produce a greater error in determining g than a small error in d? Ans: The values of t are squared, whereas in calculating g the values of d are not.

How would an increase in friction between the glider and the air track affect the calculated value of g? Ans: The value of g would be less, because the ratio of d to t would be less, producing a lower value for the acceleration measured and thus a lower experimental value for g.

Calculate the acceleration down your incline if it were placed on the surface of the moon. The acceleration due to gravity on the moon is one-sixth that on Earth. Ans: Using the example values for h and L, h = 30 cm and L = 300 cm a=

gh = L

980 × 30 cm 6

300 cm

= 16.3 cm s 2 .

The force of gravity on an object of mass m is given by the following: F=

GmM E

b R + hg

= mg

where M E and RE are the mass and radius of Earth, respectively. How far from Earth must the mass m be positioned for F to equal zero? Ans: The value of F will never go to zero but it approaches zero as RE + h approaches infinity.

Experiment 6

Newton’s Second Law INTRODUCTION The relationship among mass, acceleration, and unbalanced force is basic to the understanding of physical science, and this experiment provides students with a method for grasping the meaning of these relationships. They learn firsthand that F in the equation F = ma is the unbalanced or net force, that m is the total mass being moved by the unbalanced force, and that a is the acceleration of the total mass in the direction of the unbalanced force. For a good comparison of results between the two values of acceleration obtained from the data taken in the experiment, use a precision ball-bearing pulley (Cenco 75655 or equivalent), and make sure that it is mounted and rotates in a true vertical plane. This reduces frictional forces to a minimum. Also, all masses should be checked on a balance because the values marked on them may be incorrect. A small error in one of the masses can produce a large error in the value of the unbalanced force. The problems caused by the string bouncing out of the pulley groove and the masses falling down when the downward moving mass hits the floor (especially when students use excessive mass for the rider) can be solved by using a different apparatus obtainable at little or no cost. Take two aluminum soft drink cans and remove the top lids. Put two small holes opposite each other in the sides of the cans near the open top, through which the string can be attached. Place a small amount of soft sponge in the bottom of each can. After the total mass of each can plus the sponge is determined with a balance, the cans may be used to carry any available masses in the laboratory. The following data were obtained using paired 500-g and 1000-g brass masses. Slotted masses, which were placed on top of m1 and/or m2 , were used as riders. The same string was used for obtaining all data. Only the average times are given in the following tables. The distance (d) can be adjusted for any chosen value by varying the height of the pulley above the floor. A drop distance of at least 1 m is recommended in order to obtain good time measurements.

SAMPLE DATA AND CALCULATIONS: DATA TABLE 6.1 Trial 1

Trial 2

1.20 m

1.23 m

Unbalanced force (F), in N

0.020 × g*

Total mass (M)

2.020 kg

1.020 kg

Average time (t)

5.06 s

3.69 s

Distance (d)

The acceleration due to gravity, 9.80 m s .

DATA TABLE 6.2 Trial 1

Trial 2

Acceleration from measurements of distance and time

0.0937 m s2

0.181 m s2

Acceleration from Newton’s second law

0.0970 m s2

0.192 m s2

3.46%

5.90%

Percent difference DATA TABLE 6.3 Trial 1

Trial 2

1.20 m

Unbalanced force (F), in N

0.040 × g*

0.020 × g*

Total mass (M)

2.040 kg

2.020 kg

Average time (t)

3.54 s

5.06 s

Distance (d)

The acceleration due to gravity, 9.80 m s .

DATA TABLE 6.4 Trial 1

Trial 2

Acceleration from measurements of distance and time

0.192 m s2

0.0937 m s2

Acceleration from Newton’s second law

0.194 m s2

0.0970 m s2

1.04%

3.46%

Percent difference

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

What have the data and calculations shown concerning the acceleration, first, as a function of the mass when the unbalanced mass is held constant, and second, as a function of the unbalanced force when the total mass is held constant? Answer this by completing the following two statements: (a)

When the total mass that is accelerating increases (the unbalanced force is held constant), the acceleration will __________________________________________ .

(b)

When the unbalanced force increases (the total mass is held constant), the acceleration will _______________________________________________________________ . (a) decrease (b) increase

Ans: 2.

Distinguish between mass and weight, i.e., define both and explain the difference. Ans: Mass is a quantity of matter and a measurement of the amount of inertia that an object possesses. Weight is the force of gravity exerted on any object near Earth’s surface.

Should the mass of the string be added to the total mass being moved by the unbalanced force? Why or why not? Ans: Yes, because the string is part of the total mass being moved by the unbalanced force but if a light string is used it can usually be neglected.

Since it was the addition of m3 to the system that produces the unbalanced force on the system in Procedure 1, why is the unbalanced force in Procedure 2 equal to g m2 + m3 − m1 instead of simply m3 g ? m1 and m2 are supposed to be the same value, but this is not always true. There may Ans: be a slight difference from the values stamped on them.

Explain where at least three (3) possible sources of error could have occurred in this experiment. Ans: Possible sources of error are: inaccurate time measurements, having m1 and m2 not exactly equal to each other, poor measurements of the height (d), friction in the pulley system, badly calibrated riders, or string heavy enough to effect the results.

Which of the sources of error that you cited in question 5 do you think most affected your experiment? Why? Ans: The greatest possible source of error is probably in the measurement of the time because of problems in determining the moment of starting the timer and the exact moment the mass hits the floor. Also individual reaction times usually cause rather large errors.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

How would the data be influenced by an increase of friction in the pulley? Ans: All measured accelerations would have lower values. 25

In the experiment, m1 and m2 have the same value. How would the data change if m1 and m2 were slightly different in value? Ans: The values for the unbalanced force would be slightly more or less. These values would not be known to the observer, and the data collected would be inaccurate.

The unbalanced force acting on an object is zero. What is the direction and magnitude of the object’s velocity and acceleration? Ans: The velocity may be zero, or the velocity may have any direction and magnitude, but the velocity must be constant. The acceleration must be zero.

The unbalanced force on an object is zero. Does this imply that the object is at rest? Ans: No, the object may be at rest or in uniform motion, it just must have an acceleration equal to zero.

Experiment 7

Hooke’s Law for a Vibrating Spring INTRODUCTION Elasticity is a physical property of a material that exhibits a change in shape by the application of an external force and then returns to its original shape when the external force is removed. Elasticity is apparent in many solids used every day, from rubber bands and springs to railroad rails. The proportionality between force and change in shape is known as Hooke’s law. We use the term law here, but the relationship stated is an empirical observation, not a law in the strictest sense. A vibrating mass executes simple harmonic motion (SHM) when the restoring force is at all times proportional to the displacement. A good example of a system undergoing SHM is a mass on a vibrating spring. In this case, the key to interpreting the motion is Hooke’s law. The law can be stated in equation form as F = − k x where the minus sign indicates that x and F are in opposite directions. The period of a vibrating spring is given by the equation T = 2π

M k

The common factor here is the spring constant k, whose value is first determined experimentally using Hooke’s law and then inserted into the equation for the period of vibration. The period can then be compared with the actual vibration period when this quantity is measured using a simple clock or stopwatch. Since it is difficult to measure the period of a single vibration, the student should measure the time for 20 oscillations and divide by 20 to get the period of the vibrating system. This method should allow for at least three significant figures to be carried through the calculations on this experiment. If time is pressing, the number of repetitions in each timing run can be reduced to 10 or even 5, but the accuracy of the calculations will decrease accordingly.

EQUIPMENT AND PROCEDURAL TIPS 1.

Springs can be found in most local hardware stores. Conventional “jolly balance” springs can be used if they are already available, but they can be easily stretched out of shape if too much weight is added by the beginning student, and they are quite expensive. If you buy springs locally, take along a 100-g mass and a ruler, and choose a style that allows about 2 to 3 cm of elongation when the 100-g mass is hung from it. Such a spring should cost from $2.00 to $3.00 each. 27

Have students take all height measurements from the same point at or near the bottom of the hanging spring. If the bottom of the suspended mass is used, its position may vary when different masses (of different sizes) are hung from the spring. This will, of course, not be a problem if a long 50-g weight hanger is used and all masses are placed on top of its weight platform.

Time measurements are very important and should be taken with as much accuracy as your clock or stopwatch allows. It is possible to use a standard sweep second-hand on a watch to make the time measurements, but more vibration cycles may need to be counted, especially at low mass loads.

Make sure that students count the correct number of complete oscillations. When the timing cycle is begun, they should say “start” and then “one, two,” etc. after each successive vibration. If they begin the timing on the count of “one” and count to “twenty,” they will have timed only 19 actual oscillations.

The support stand and clamp must be fairly rigid. If they rock or bend while the spring system is oscillating, the experimental results will not be accurate.

The data in the following tables are for a spring purchased at a local hardware store. DATA TABLE 7.1 (Record units with all data values.) Starting height h0 = ___________ Height Above the Table h

X = h0 − h

Force F = mg

Trial

Total Mass on Spring m

0.100 kg

0.654 m

0.043 m

0.980 N

22.6 N m

0.200 kg

0.610 m

0.087 m

1.96 N

22.5 N m

0.300 kg

0.575 m

0.132 m

2.94 N

22.3 N m

0.400 kg

0.526 m

0.171 m

3.92 N

22.9 N m

0.500 kg

0.481 m

0.216 m

4.90 N

22.7 N m

0.600 kg

0.436 m

0.261 m

5.88 N

22.5 N m

Average value of k for all six trials

F X

22.6 N m

Note: The incremental mass is 0.100 kg, and the starting height h0 = 0.697 m . The spring mass = 0.0295 kg.

DATA TABLE 7.2 (Record units with all data values.) Theoretical Effective Measured Experimental Percent Error in Mass (M) of Period (T) Time (t) for 20 Period, Time Experimental Spring and Oscillations for 1 Vibration Compared to M T = 2 π Hanging Theoretical t k T= Trial Number Mass Periods 20 1

0.110 kg

0.438 s

09.10 s

0.455 s

3.9%

0.210 kg

0.605 s

12.80 s

0.640 s

5.7%

0.310 kg

0.736 s

15.70 s

0.785 s

6.7%

0.410 kg

0.846 s

17.90 s

0.895 s

4.9%

0.510 kg

0.943 s

20.2 0s

1.010 s

7.1%

0.610 kg

1.030 s

21.80 s

1.090 s

5.8%

ANSWERS TO QUESTIONS IN THE LAB MANUAL

b g

Why must you use the same reference point h0 on the spring for each subsequent height measurement? Ans: The reference point for measuring the height above the table must be chosen in such a way that elongation of the spring is the variable that is consistently measured. This measurement may not be consistent if the masses vary in size or if the bottoms of the masses are used as the height reference point.

Why did you use the average value for k from Data Table 7.1 rather than one of the individual trial values? Ans: When any data are taken, the average value of several measurements is a better indication of that measurement than any single value that was taken.

Do a calculation of the theoretical value of the period for the oscillating mass for Trial 4 without correcting the effective mass by adding one-third of the mass of the spring. Again determine the percentage error between the experimental and theoretical values of the period of oscillation. Do you agree that the effective mass must be corrected in this way to obtain good results? M T = 2π Ans: k If M = 0.400 kg instead of the corrected value of 0.410 kg, T = 2π

b g 0.0177 s

0.400 kg = 2 314 . 22.6 N m

. T = 6.28 0133 = 0.835 s

0.846 s − 0.835 s × 100 = 13% . 0.846 s In this case the correction is quite small, but it should be made anyway. For the 0.100-kg mass case (Trial 1), the error is about 4.6 percent, and using the corrected mass is much more important.

The error introduced is

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What is simple harmonic motion? Ans: Simple harmonic motion is a vibration caused by a force that follows Hooke’s law. It is motion in which the acceleration is directed toward the center of the motion and is directly proportional to the displacement from its equilibrium position.

Explain two methods that could be used to find the spring constant of a spring experimentally. Ans: (a) Measure the elongation caused by hanging various masses from the spring, and use Hooke’s law to calculate k. (b) Measure the period of oscillation for a vibrating mass hung on the spring, and find k from the equation for the period of a simple harmonic oscillator.

How is the mass hung on the spring related to the force that actually stretches the spring? Ans: The force that Earth exerts on the hanging mass is called weight. The weight of the mass is the actual force that stretches the spring. Weight = mg, where g is the acceleration of gravity, 9.8 m s2 .

Why must you be careful in how you start the timing when you measure the time increments for multiple oscillations of a vibrating system? Ans: If you begin the timing on the count of “one” and count to “twenty,” you will have timed only 19 actual oscillations.

What would happen if you hung so much mass on the spring that it exceeded the spring’s elastic limit? Ans: The spring would stretch out of shape and become permanently deformed, making it unsuitable for further use.

Experiment 8

Centripetal Acceleration and Force INTRODUCTION A unique feature in this experiment is the apparatus. This experiment is completed in many laboratories with a centripetal force apparatus costing hundreds of dollars. Here the apparatus is constructed from low-cost materials found in most laboratories and shops. The crossbar attached to the rotating rod should be of equal length on each side of the rotating rod to counterbalance the rotating system. Two students should work together when data are taken. One student will pull the string to produce the rotation, while the second student will obtain the time for 20 or more revolutions. There should be a minimum of 10 s for the number of revolutions counted. A couple of graphs can be included in this experiment. They have been omitted here because students are often tired of plotting graphs by the time this experiment is assigned. The two alternative or additional questions can be asked in place of graphing. The two graphs are (1) Fc versus v and (2) Fc versus v2.

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

Assume that the string breaks while the mass (m) was in uniform circular motion. If the spring is also suddenly unhooked from the mass at the same instant that the string breaks, what will happen to the mass? Ans: The mass would fly off tangent to the circle of rotation at the instant the spring becomes unfastened.

How did Fc change when you redid the experiment using the larger mass? Did you expect this? Why or why not? Ans: The centripetal force increases with an increase in mass. This is expected, because the centripetal force is directly proportional to the mass as shown in Eq. 8.2 in the Laboratory Guide.

Why must the string held in the student’s hand be pulled at a uniform rate? Ans: For the mass on the plumb bob to pass directly above the masking tape marker on each rotation its angular velocity must be constant and this can only happen if the string is pulled at a uniform rate.

Calculate the centripetal acceleration for each of the two trials. Ans: This answer will of course vary with each experimental setup so no exact answers are given here. Use Eq. 8.2 to make these calculations.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What type of graph would be obtained when the centripetal force is plotted on the y axis and the velocity is plotted on the x axis? Ans: A parabola

What type of graph would be obtained when the centripetal force is plotted on the y axis and the velocity squared is plotted on the x axis? Ans: A straight line

Is it possible for an object to have nonzero acceleration but still have the magnitude of its velocity be zero? Ans: Yes, an object moving in a circle at constant speed is still changing its direction of travel and thus has a centripetal acceleration.

Experiment 9

Laws of Equilibrium INTRODUCTION Statics is the study of objects undergoing zero acceleration under the action of two or more forces. When this is true, the object is in equilibrium. This experiment is a study of static equilibrium. Procedure 1 studies the conditions of equilibrium of concurrent coplaner forces. Procedure 2 studies non-current coplaner forces. The experiment provides students with a direct method for observing torques and the rotations they can produce before the system is brought into equilibrium. Using the moment-of-force apparatus, students observe and measure the acting forces and moment arms. The instructor should give a brief lecture on parallel forces, introducing the concept of equilibrant force. When two parallel forces are placed in equilibrium by a third force, applied in the experiment with the moment-of-force apparatus at the center of gravity, the clockwise and counterclockwise torques are equal. From this, students should learn that the two forces are inversely proportional to the lengths of the moment arms. The following three tables give typical data for the experiment.

DATA TABLE 9.1 Mass of the bar 0.1195 kg Point of balance of the bar 0.502 m (Should be near the 50 cm point)

Part

Force (N)

Lever Arm (m)

Clockwise Torque (N m)

Counter-clockwise Torque (N m)

0.98

0.202

—

–0.198

0.49

0.406

+0.199

—

0.98

0.400

—

–0.392

Percent Difference in Torques 0.5%

1.02%

1.96

0.202

+0.396

—

0.98

0.400

—

–0.392

0.49

0.200

—

–0.098

1.96

0.250

+0.490

—

*Note: Table in Laboratory Guide is incorrect because it does not have enough horizontal rows. Have students add rows or make up new table.

Up Forces (N) Symbol

DATA TABLE 9.2 Center of Torques at Zero End of Bar Lever Arm Counter-clockwise Down Forces of Force Torques (N) (m) (N m)

Clockwise Torques (N m)

+2.40

—

0.18

–0.43

—

+1.90

—

0.82

–1.56

—

–1.00

0.06

—

+0.06

—

–2.00

0.64

—

+1.28

—

–1.46

0.50

—

+0.73

Sum

+4.30

–4.46

–1.99

+2.07

Up Forces (N) Symbol

DATA TABLE 9.3 Center of Torques at Zero End of Bar Lever Arm Counter-clockwise Down Forces of Force Torques (N) (m) (N m)

Clockwise Torques (N m)

5.80

—

0.26

—

+1.51

3.60

—

0.66

—

+2.38

—

4.90

0.19

–0.93

—

FX*

0.84

0.84 FX

—

1.20

0.50

–0.60

—

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

Explain the difference between force and torque. Ans: A force is any quantity capable of producing motion. A torque is a force times the perpendicular distance from the line of action of the force to the axis of rotation that is capable of producing angular motion.

State the two laws of equilibrium. Ans: An object is in equilibrium when (1) the sum of all forces acting on the object is zero and (2) the sum of all torques acting on the object is zero.

How does the fact that the center of gravity of the meter stick, when it is balanced, may not be exactly at the 50-cm point affect your measurements? Ans: All length measurements will not be correct unless a correction is made. This, in turn, will give inaccurately calculated torques if the correction is not properly made.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

When equilibrium was established in Procedure 1, Part 1, what was the relationship between the two forces and the length of their lever arms? Ans: They were each inversely proportional to one another.

Is it possible for a moving object to be in equilibrium? Ans: Yes. If the sum of all forces and torques add to zero, the object will be in equilibrium even though it may be moving at a constant velocity.

How do you determine if a particular torque should be considered + (positive) or – (negative)? Ans: By convention a torque that could produce a counter-clockwise rotation is + (positive), a torque that could produce a clockwise rotation is – (negative).

In this experiment which forces are considered + (positive) and which are – (negative)? Ans: Since all relevant forces are in the y-plane, by convention all upward forces are + (positive) and all downward forces are – (negative).

Experiment 10

Principle of Work Using an Inclined Plane and Pulleys INTRODUCTION The principle of energy conservation states that the energy of an isolated system remains the same while forces, velocities, and other physical constants change. Energy and work are measured in the same units. In fact, energy is defined as the ability or capability of a body to do work. We define work done by a constant force to be equal to the component of the force parallel to the displacement times the displacement. The inclined plane and pulley are machines used in this experiment to study the principle of work. Before beginning the experiment, the instructor should emphasize that a machine is a force changer that multiplies forces at the expense of displacement but does not multiply work. Stress that (1) the work out is always less than the work in, (2) the ratio of the output (work) to the input (work) gives the efficiency of the machine, and (3) this ratio is always less than unity. The following tables give typical data for the experiment.

W Weight of Car and Added load Angle of (if any) Plane (N)

DATA TABLE 10.1 Note: Weight = mass × g D H F Percent (from Difference spring Between balance) Applied Theoretical Actual and Actual Force F W D Theoretical M.A. = M.A. = M.A. (m) (m) (N) F H

25°

3.72

0.98

3.80

0.25 m

0.065 m

4.98

27%

25°

7.63

1.96

3.89

0.25 m

0.065 m

4.98

25%

45°

3.72

2.20

1.69

0.25 m

0.130 m

1.92

13%

45°

7.62

4.66

1.64

0.25 m

0.130 m

1.92

16%

DATA TABLE 10.2 Percent Theoretical Difference D H (D2 – D1) (H2 – H1) Between D M.A. = (m) (m) H Actual and Theoretical M.A.

Load L (N)

Force F (N)

Actual L M.A. = F

7 strands

9.8

1.5

6.53

0.28

0.04

7.00

06.9%

5 strands

9.8

2.2

4.45

0.30

0.06

5.17

15%

EFFICIENCY OF MACHINES The following are the calculated efficiencies of these machines. Inclined plane 15° (Trial 1) Inclined plane 15° (Trial 2) Inclined plane 45° (Trial 1) Inclined plane 45° (Trial 2) 7-String pulley system 5-String pulley system

% Efficiency = % Efficiency = % Efficiency = % Efficiency = % Efficiency = % Efficiency =

76.3% 78.1% 88.0% 85.4% 93.3% 86.1%

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

Define work in terms of applied force and distance. Ans: The work done by a constant force acting on an object is the product of the magnitude of the force (or component of force) and the parallel distance through which the object moves while the force is applied.

Why must the spring balance be moving slowly at a constant speed when the applied force is measured? Ans: The applied force must not vary in value and the acceleration must be zero, so the speed must be constant.

Distinguish between TMA and AMA. Ans: TMA is the ratio of the distance the input force moves to the distance the output force moves. AMA is the ratio of the output force to the input force of any machine.

Is the TMA always greater than the AMA for a machine? Explain. Ans: TMA is always greater than AMA because friction is always present and must be overcome by the input force, although this effect may be so small that it is difficult to detect, especially at slow speeds.

Define efficiency. Give an example. Ans: Efficiency is the ratio of the work output to the work input. Example: The work input to a machine is 50 N ⋅ m and the work output is 40 N ⋅ m . The efficiency is 40 50 , or 80 percent.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

State how the mechanical advantages and the efficiency of the inclined plane vary with the inclination of the plane. Ans: Both AMA and TMA decrease with an increase in the angle theta θ . The efficiency will remain the same if friction is small enough to be neglected.

Which is the best indication of the usefulness of a machine, the actual mechanical advantage or the theoretical mechanical advantage? Ans: Since machines are generally used to increase the force that can be applied to a load, the actual mechanical advantage is usually the most useful. The theoretical mechanical advantage is sometimes useful, however, when the machine in question is judged by how far the load is moved during its operation.

Experiment 11

Waves INTRODUCTION Students have fun doing this experiment. They tend to be noisy, and at the beginning of the laboratory period the instructor should caution them to obtain the data with the least noise possible. In Procedure 5 there will be interference between other nearby resonance columns. Organize the class so that only one or two resonance columns are operating at the same time, preferably at opposite ends of the laboratory. Allow some students to start with Step 5 in the Procedure. This will minimize the interference. Caution: If the resonance apparatus has a glass tube (some have plastic tubes), be sure to warn students not to touch the tube with a vibrating tuning fork. The vibrating fork may break the glass. Also, do not allow students to empty the water from the resonance apparatus. They may be careless and break the glass tube. Experimental data for this experiment are mostly observational, and no sample answers or data are provided here.

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

Define wave motion, and give an example. Ans: Wave motion is the emanation of energy from the disturbance of matter. Light and heat energy radiating from a burning match are examples.

Distinguish between longitudinal and transverse waves. Ans: A longitudinal wave is a wave in which the particle displacement is in the same direction as the wave motion. A transverse wave is a wave in which the particle displacement is perpendicular to the wave motion.

Define the velocity, wavelength, and frequency of a wave and give an example of each. Use symbol notation in the definitions. Ans: The velocity (v) of a wave is the distance the wave travels per unit of time. The velocity of a sound wave in air is 330 m/s at 0°C. The wavelength (λ) is the distance between two similar points on any two consecutive waves. The wavelength of the sound wave is given in meters. The frequency is the number of cycles per unit time. The frequency (f) of the sound wave is measured in Hz.

State the relationship between velocity, wavelength, and frequency of a wave. Ans: Velocity = frequency × wavelength.

Define the amplitude of a wave, and give an example. Ans: Amplitude is the maximum displacement of a wave from its equilibrium position. An example is the maximum displacement of the vibrating string which moves 0.100 meters above and below its equilibrium position. The amplitude in this case is 0.100 meters.

State the relationship between the frequency and the period of a wave. Ans: frequency = 1/period.

How do your measurements in Procedure 2 predict the way that the velocity of a wave in a rubber cord varies with the tension in the cord? Ans: The wave velocity is proportional to the tension in the cord so wave velocity increases as the tension is increased.

Why is there a maximum vibration of the air molecules at the open end of a resonance tube? Ans: The air molecules can vibrate more freely at the open end of the tube allowing them to achieve maximum displacement.

How does the velocity of sound in air vary with the air temperature? Ans: The velocity of sound in air increases with an increase in the air temperature by about 0.6 m/s for each Celsius degree the temperature rises.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What effect does increasing the amplitude of the vibrations have on the data? Ans: Neither velocity, frequency, nor wavelength are functions of the amplitude.

With a drawing similar to Figure 11.6 in the Laboratory Guide, illustrate the particle displacement in a resonant tube open at both ends. Ans: Refer to this drawing in the Laboratory Guide, your drawing here should show both ends of the tube open (as the upper part of Figure 11.6 shows), so the particle displacement will be a maximum at both ends.

Experiment 12

Interference of Light Waves INTRODUCTION The Keuffel and Esser wave pattern transparencies may be purchased from most local bookstores. If the store does not have them in stock, they can order them for you from Keuffel and Esser. Most students can complete this experiment in 1 hour or less; therefore, you may want to do it in conjunction with Experiment 27 on spectroscopy or with a film on wave motion and/or with a film on interference of light waves. See the Instructor’s Guide for the textbook for film list and suppliers. The instructor should check each student’s drawing at the beginning of Procedure 1. A wrong start will produce a bad drawing and poor data. The following are typical data for the experiment. The values for x and y will depend on how far from S1 and S2 the student places the vertical line that is drawn perpendicular to the line labeled n=0.

d (cm)

DATA TABLE 12.1 y (cm)

y1 = 2.1

x1 = 18.0

0.23

y2 = 4.2

x2 = 18.3

0.23

y3 = 6.8

x3 = 19.0

0.24

y1 = 1.5

x1 = 17.8

0.25

y2 = 2.9

x2 = 18.0

0.24

y3 = 4.5

x3 = 18.3

0.25

x (cm)

λ c Calculated (cm)

Average value for λ c

0.24

Measured value for 20 wavelengths = 4.8 cm 1 wavelength = 0.24 cm

ANSWERS TO QUESTIONS IN THE LAB MANUAL

Define and explain phase as it relates to wave motion. Ans: The phase of the vibrating particle experiencing the disturbance is a measure of its position (an angular measurement) in reference to its equilibrium position.

What would you observe when two light waves meet in phase with one another? Ans: Two waves are in phase with one another when they are displaced at all times in the same direction will produce a brighter region on a screen.

When referring to wave motion, what is the meaning of the expression out of phase? Ans: Out of phase refers to the position of the vibrating particle’s displacement from its equilibrium position relative to another particle’s displacement when two waves meet out of phase the crests of one wave coincide with the troughs of the other.

What conditions must be met for complete destructive interference between two waves? Ans: For complete destructive interference between two waves, they must be 180° out of phase at all times. The two waves must also have the same amplitude, the same wavelength, and be traveling in the same direction.

Distinguish between the terms slit width and slit separation. Ans: Slit width refers to the width of the opening, and slit separation refers to the distance between two separate slits.

How does the separation of maxima on the screen vary with slit separation? Ans: The separation of maxima on the screen is inversely proportional to slit separation.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

The relationship among frequency, wavelength, and velocity of a wave is given by the equation v = f λ . Calculate the wavelength of green light if the frequency is 5.7 × 1014 Hz . The velocity of light is 3 × 1010 cm s . v 3 × 1010 cm s ° Ans: λ= = = 0.5263 × 10 −4 cm = 5263 A f 5.7 × 1014 cycles s

Which regions on the screen show constructive interference and which show destructive interference if monochromatic light is shown through a double slit? Ans: The light regions occur where constructive interference takes place, and the dark regions coincide with destructive interference.

Experiment 13

Plane Mirrors and Index of Refraction of Light INTRODUCTION Most students enjoy this experiment because it is their first experience with seeing the relationship between object and image characteristics of mirrors and lenses. If the laboratory can be darkened, the candle makes a good object because it produces an image that is pretty and has motion. Caution students about the danger of burning their shirt or dress sleeves. If the laboratory cannot be darkened, then a flashlight can be used as a light source. Place a small black arrow on a piece of cheesecloth and put it over the face of the flashlight by removing the ring that holds the glass plate in place. SAMPLE DATA

Trial

θi

DATA TABLE 13.1 Estimated Error

22.2°

± 0.4°

21.9°

± 0.4°

37.0°

± 0.4°

36.8°

± 0.4°

49.3°

± 0.4°

49.5°

± 0.4°

Trial

DATA TABLE 13.2 Object Distance (AM), in meters

θr

Estimated Error

Image Distance (BM), in meters

0.123

0.125

0.171

0.169

DATA TABLE 13.3 Angle of Incidence

θr

sinθ r

Index of Refraction sinθ i n2 = sinθ r

0.5764

22.3°

0.3792

1.52

0.6560

25.4°

0.4288

1.53

θi

sinθ i

Left (1)

35.2°

Right (2)

41.0°

Surface

Angle of Refraction

Calculate the percent difference between your two values for the index of refraction. % difference = 0.66%

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

State the law of reflection, and give an example showing its relevance to this experiment. Ans: The law of reflection states that when the reflection of light is from a plane specular surface, the angle of reflection is equal to the angle of incidence, and the incident ray, the reflected ray, and the normal to the surface at the point of contact are all in the same plane.

Distinguish between reflection and refraction. Ans: Reflection is a change in wave direction due to the bouncing back from a boundary between two media. Refraction is a change in wave direction due to a change in the velocity of the wave as it moves into a new media.

State the law of refraction (Snell’s law), and explain how it can be used to find the index of refraction for a glass plate. Ans: Snell’s law states that for any transparent substance, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is the same for all angles of incidence and is equal to the inverse ratio of the refraction indexes of the two materials. m2 sin θ i . = m1 sin θ r

If the index of refraction for a glass plate is higher than the index of refraction for water, which medium will bend a light beam more when the light beam enters the new material from the air? Explain your choice. Ans: The glass, because glass has a higher index of refraction.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

When studying the angle of incidence in the refraction process, how must this angle be measured? Ans: When light rays strike a boundary, the angle of incidence of the incoming ray is measured in respect to the normal line drawn to the interface surface.

Describe the image formed by an object located 1 m in front of a plane mirror. Ans: The image will be virtual, erect, 1 m behind the mirror, and the same size as the original object.

When a light beam enters glass from air the beam is refracted. Why does this happen? (Hint: What physical property of the light changes?) Ans: Light is bent on entering glass from air because the velocity of the light is less in the glass than it is in air.

Is it possible to form a real image with a plane mirror? Why or why not? Ans: No, because the focal length of a plane mirror is infinite and the image is always behind the mirror’s surface.

Which color was deviated most by the prism used in the experiment? Ans: Violet

Experiment 14

Mirrors, Lenses, and Prisms INTRODUCTION If the laboratory does not have good mirror and lens holders, caution the students to be careful with makeshift holders. Also, point out the positions of the mirror and the screen as shown in Figure 14.3 in the Laboratory Guide. Many students place the screen in front of the light source, thus blocking the rays of light to the mirror. If the sun is visible from the laboratory window or students can step outside to see it, they will obtain a more accurate measurement of the focal length by using it as a source of distant light. The sun is also the best source of white light for Procedure 1, Step 5.

Figure 14.3 Top view of optical bench. Model ray diagrams are explained with drawings for both the concave mirror and the convex lens in the Laboratory Guide. The following data are typical for a concave mirror with a focal length of 20 cm and a double convex lens with a focal length of 15 cm.

Trial

DATA TABLE 14.1 Data for a 20 cm focal length mirror. Focal Length of Mirror

19.8 cm

19.9 cm

20.3 cm

Average

20.0 cm

Position of Light Source

DATA TABLE 14.2 Data Table for Mirror Data for a 20 cm focal length mirror. Information on Image Position of Real or Erect or Image Virtual Inverted

Larger or Smaller

+ Infinity (at least 8f)

20.1 cm

Real

Inverted

Smaller

At 4f

27.0 cm

Real

Inverted

Smaller

At 2f = R

40.0 cm

Real

Inverted

Same size

Between R and f

30.0 cm

Real

Inverted

Larger

At f

+ Infinity

Real

Inverted

Larger

behind mirror

Virtual

Erect

Larger

Between f and mirror

Position of Light Source

DATA TABLE 14.3 Data Table for 15 cm Length Lens Information on Image Position of Real or Erect or Image Virtual Inverted

Larger or Smaller

+ Infinity (at least 8f)

15.0 cm

Real

Inverted

Smaller

At 4f

20.0 cm

Real

Inverted

Smaller

At 2f

30.0 cm

Real

Inverted

Same size

Between 2f and f

50.0 cm

Real

Inverted

Larger

At f

+ Infinity

Real

Inverted

Larger

Same size of lens as object

Virtual

Erect

Larger

Between f and lens

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

Describe the difference between a real and a virtual image. Ans: A real image can be focused on a screen, whereas a virtual image cannot. The observer must look into a lens or mirror to see a virtual image.

Where must the object be located with respect to a concave mirror and its focal point to produce a real image? Ans: The object must be located at a distance greater than the focal length of the mirror, in front of a concave mirror, in order to produce a real image.

Where must the object be located with respect to a converging lens and its focal point to produce a virtual image? Ans: The object must be located between the converging lens and its focal point to produce a virtual image.

At what distance must the object be placed on the principle axes in front of a concave mirror for the image to appear the same size as the object? Draw a ray diagram to illustrate your answer. Ans: The object must be placed at the radius of curvature.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Where must an object be placed to form an erect image with a concave mirror? Ans: The object must be placed inside the focal point.

Describe the image formed by a convex lens when the object is closer to the lens than its focal length. Ans: The image will be virtual, erect, and larger.

What type of images are formed by a concave (diverging) lens? Ans: Only virtual images are formed by a concave (diverging) lens.

What type of mirror is needed to form real images? Ans: Real images can be formed by a converging mirror.

Under what conditions will a converging (convex) lens produce a real image? Ans: A converging lens forms a real image when the object distance is greater than the focal length.

Experiment 15

The Refracting Telescope INTRODUCTION Caution students to take special care in placing the lens in the lens holder and mounting the lens holder with lens on the meter stick. Make sure the fittings are tight; otherwise, the lens will fall out and be chipped or broken. Also, the objective lens and the eyepiece lens must be in line and sharing a common optical axis. The lenses must be mounted perpendicular to the optical axis. Students should have had enough experience drawing ray diagrams to complete Alternative Question 2 without help from the instructor. Some students will need help with number 4 in the procedure. Even with help, others may not be able to see through the telescope to determine magnification. Let these students make the necessary marks on the blackboard. A good choice is to use the double convex 15 cm focal length lens used in Experiment 14 for the objective lens and a double convex 5 cm focal length lens for the eyepiece. This will give a refracting telescope with a magnification of three. The data for Table 15.1 is for recording the focal lengths for the two lenses and the distance of separation when the telescope is constructed. This distance is slightly less than the sum of the two focal lengths.

SAMPLE DATA DATA TABLE 15.1 Trial 1 1.

Trial 3

Position of objective lens on meter stick

10.0 cm

12.0 cm

15.0 cm

Position of screen on meter stick

25.0 cm

27.1 cm

29.9 cm

Focal length of objective

15.0 cm

15.1 cm

14.9 cm

Average focal length of objective fo

a f

15.1 cm

Position of eyepiece lens on meter stick

10.0 cm

12.0 cm

15.0 cm

Position of screen on meter stick

15.1 cm

16.8 cm

20.0 cm

Focal length of eyepiece

5.1 cm

4.8 cm

5.0 cm

a f

Trial 2

Average focal length of eyepiece fe

4.96 cm

Position of objective lens on meter stick

10.0 cm

15.0 cm

20.0 cm

Position of eyepiece on meter stick

30.4 cm

35.6 cm

40.0 cm

Distance between lenses

20.4 cm

20.6 cm

20.0 cm

Average distance between lenses (l)

20.3 cm

(a)

Vertical height of image hi

9.2 cm

(b)

Vertical height of object ho

3.1 cm

SAMPLE CALCULATIONS 1.

Sum of average focal lengths of lenses (from 1 and 2)

L = f o + f e = 20.1 cm

Actual Distance between lenses (from 3 in Data Table 15.1)

20.3 cm

% difference =

0.99 %

Percentage difference between L and ! 2.

height of image hi = height of object ho

(a)

Magnification (from 4) =

(b)

Magnification (from 1 and 2) =

(c)

Percentage difference between magnification values (a) and (b) = 1.67 %

focal length of objective f o = focal length of objective f e

3.00 cm 3.05 cm

Diameter of objective lens = 4.00 cm (measured in lab) Light-gathering power of your telescope = 39.1

12.7 cm diameter of the objective lens in cm Calculate the resolving power of the telescope constructed in the laboratory. Show your work. Resolving power of your telescope = 3.18 Resolving power =

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

What is the simplest way to increase the magnifying power of a two-lens telescope? Hint: The eyepiece lens is often mounted in a short tube that can easily be removed from the main part of the telescope. Ans: The magnifying power of the telescope is equal to f o f e . Therefore, increasing f o or decreasing f e will increase the magnifying power. Usually the eyepiece is changed out so that a shorter focal length one is inserted.

Why is it better if a distant object outside the window of the laboratory is used when the focal lengths of the two lenses were determined? Ans: To obtain parallel rays of light so the focal length determination will be more accurate.

Calculate the light-gathering power of the 508 cm (200 in) telescope on Mount Wilson. Show your work. Ans: The accepted value for the diameter of the lens opening of the average human eye is 0.64 cm so; the light-gathering power =

( diameter of object lens )2 ( 508 cm )2 = = 631,000 . ( diameter of len of eye )2 ( 0.64 cm )2

Calculate the resolving power of the 508-cm (200-in) telescope on Mount Wilson. Show your work. 12.7 cm 12.7 cm Ans: Resolving power = = = 0.025 . diameter of objective lens in cm 508 cm

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

The two-lens astronomical telescope constructed for the experiment produces an inverted image of the object. To convert the astronomical telescope to a terrestrial telescope, mount an inverting lens in the system. Determine the position for an inverting lens that will not change the magnification of the telescope. Ans: Place the inverting lens so that the image from the objective lens is at the radius of curvature of the inverting lens. 55

What are the functions of the objective lens in the refracting telescope? Ans: The objective lens collects light from the distant object and brings the light rays to a focus, forming an image that is the object for the eyepiece.

Draw a ray diagram illustrating the position of the distant object and the position and size of the two images formed by the refracting telescope. Ans: See figure below.

Experiment 16

Color INTRODUCTION This experiment is designed to use a minimum of equipment while showing a wide variety of color phenomena. For this reason, a system of six stations has been set up, one for each procedure, so that the students can “rotate” through these stations during the lab period, but only one set of equipment needs to be available for each procedure. If more equipment is available, two or more stations for each procedure could be set up, but from 10 to 14 students can easily complete this experiment in 2 hours with only one station for each procedure.

EQUIPMENT AND PROCEDURE TIPS Procedure 1: Any diffraction grating and white light source can be used. Some very inexpensive, hand-held diffraction units are available. They are made out of a short plastic tube with the grating in one end and a thin single slit in the other. Another method is to provide 2 in × 2 in mounted gratings and to tape black paper over the front of a high-intensity lamp so that the white light only emerges from a narrow slit in the paper. This lighted slit can then be viewed directly through the diffraction grating. Sample Data: The hues of color seen, from long wavelength to short wavelength, are red, orange, yellow, green, blue, and violet. Note: Purple is not a spectral color. It is a combination of red and blue hues. Procedure 2: There are three methods that can be used to do this procedure. (1) 2 in × 2 in color filters can be placed in three standard 2 in × 2 in slide projectors so that each projector produces only one of the primary colors of light. These colors can then be projected and combined on a screen or white wall. This also makes a very good demonstration of the additive theory of light for use in class. (2) A Singerman apparatus can be used, which has three rheostat-controlled lamps that project colored light onto a built-in screen. These lamps can be turned on or off or adjusted in intensity to produce various color combinations. (3) The Fisher lightbox has a single lamp but comes equipped with two winglike mirrors that allow the light to be split into three beams (the primary plus two from the mirrors). A color filter can be placed in each beam, and the light from these three beams can be viewed separately or mixed together to get the desired effects. The light can be shown on a small vertical screen or if the entire apparatus is set on a white surface (or a sheet of 8.5 in × 11 in white paper), the colors can be shown and mixed on the horizontal surface.

The “white-plus-another-color” procedure can be accomplished by simply removing the primary color filter from one source to provide the white light. The white light is quite bright compared with the colored source, so you may need to provide a cardboard sheet with a small hole in the center to cut down the intensity of the white light source. The cardboard shield can be held in the beam some distance in front of the projector or mirror, thus allowing less white light to mix with the colored light. Sample Data: Primary colors of light are red, green, and blue. red + blue = magenta green + blue = cyan red + green = yellow red + green + blue = white red + white = light red (pink) blue + white = light blue (sky blue) Procedure 3: Several sources for colored plastic filters are available. You can purchase a set from a scientific supply house, or you can look around in bookstores or hobby or sewing stores for sheets of colored acetate or cellophane. If you are short on filters, the primary colors of red, green, and blue can be used, but it is best if other colors are provided. If possible, you should get a set of subtractive filters (cyan, magenta, and yellow) so that the subtractive theory of color can be studied here. These filters can be found in sets such as the one that comes with the Fisher lightbox or can be purchased separately from a scientific supply house. Local photographers also use such filters in color processing, so you may be able to borrow a set or get an old set inexpensively from them. You will want the very dark filters from these sets, which are generally not used very often anyway. The combination of two primary subtractive filters will give you blue, red, and green; all three should give black (although you probably will get a gray result unless the filters are very dark). These filters also can be combined on the surface of an overhead projector to show the secondary colors of the subtractive theory on a screen. Note: You cannot use this procedure on an overhead projector with the primary colors of light to show the additive color theory. You must have three separate light sources, each of a primary additive color. If you have a good set of primary subtractive filters, you may wish to add a procedure, 3A, in which you consider the subtractive color theory in the same way that combinations of additive filters were used in Procedure 2. Procedure 3 could then still be done using other colored filters. Students seem to enjoy making up creative names for the various colors produced by the combination of color filters. Some examples are “forest green,” “olive green,” and “apple green.” The idea here is for them to see the colors, not to name them in any special way. Sample Data: For the primary subtractive colors magenta, cyan, yellow: ! magenta + cyan = blue ! cyan + yellow = green ! yellow + magenta = red ! magenta + cyan + yellow = black (dark gray or brown)

Trial 1 2 3 4 5 6

Sample Data Using Random Color Filters Color 1 Color 2 Red Yellow Red Green Red Light blue (sky) Red Dark blue Light blue (sky) Light green (apple) Yellow Turquoise

Combination Color Orange Black Violet Purple Medium green (spruce) Olive

Procedure 4: For this procedure you will need some small cloth swatches. Get them from some old clothing, or go to a fabric store and ask for strips of several different colors. The swatches can be any size and shape but should be at least 4 in by 4 in. Try to get different textures and reflective qualities for your samples. (Fabric stores usually have some odds and ends that are quite inexpensive or even free.) Now go to a local paint store and pick out some “color chip” paint cards that you think are close to matching the cloth swatches. Another good source for these “color chips” is a home improvement center like LOWE’S or HOME DEPOT and most hardware stores also have them. Get a few extras in varying tones close to the ones you chose, to compensate for the fact that the lighting in your classroom may not be the same as in the store. Each color chip should have four to six colors on it, so you probably will have plenty of choices. (These color chips also should be free if you don’t ask for too many.) It is not essential that you have a perfect color match. Note that the questions ask for the color number that each student thinks is the best match. Often, students in the same group will disagree about the best choices. You can use these variations in color matchings to lead into a discussion of color perception and lighting variations. You can further enhance the impact of this procedure by taking two cardboard boxes and fixing them so that one interior is lighted by an incandescent bulb and the other by a fluorescent lamp. (If your room has fluorescent lights, you may need only one box with an incandescent source. However, if you do use only one box, make sure that sunlight does not “contaminate” the fluorescent light too badly. A small window of approximately 4 in by 6 in cut into the box for viewing prevents most room light from affecting your observations.) Try to find at least one cloth swatch that shows a dramatic change when viewed under the different lighting conditions. Move the swatch from one box to another to demonstrate the change in apparent color. It is best to use only one set of cloth swatches and one set of color chips and move them back and forth so that the students will not think the samples themselves are different. Sample Data: This procedure depends on your choice of cloth swatches and color chips. Color comparisons are made more accurately when the two colors being compared are close to each other. Different students may choose different color chip matches for a given cloth swatch. This variation is perfectly acceptable, since there is no absolute answer here and color perception may vary from person to person. If you perform this procedure under different lighting conditions, the results may also be quite varied. Procedure 5: Most students love to paint and mix colors. It is best to warn them in advance to wear old clothes because even water colors can sometimes stain certain fabrics. Use small bottles of tempera paint so that cross-contamination will not ruin all your paint supply. (Baby-food jars are a good size.) About two dozen inexpensive artist brushes can be provided for applying the paint, but wooden splints can be used if no brushes are available. These wooden splints or plastic spoon handles (remove the spoon part) work well because the paint can be distributed in small quantities. One or two sticks per color is enough. Have some plastic sheets or wax paper taped on cardboard about 6 in square to make palettes on which the students can carry and mix their paints. 59

Advise students to mix small quantities of paint, one or two drops at a time. Add dark colors very slowly (in small quantities) to white or you will use up a lot of white paint making tints. Add black very carefully when making tones or the color will darken too quickly. Mix the colors on the plastic palettes and then color in the circles. Several small beakers or glasses (or additional baby-food bottles) will be needed to rinse brushes between color applications. A thick layer of newspaper over all table surfaces helps in the cleanup process after the lab is over. If a sink is available, have students rinse off their plastic palettes after use so that you will not have so much clean up to do later. If the paint is placed on one table and students can work on one or two adjacent tables, several groups can do this procedure at one time. You must have more plastic palettes, brushes, and newspapers if several tables are used. When the same tone of gray is painted on white and also on black paper, the gray seems darker on the white paper and lighter on the black even though they are the same basic color to start with. This effect shows the influence of background on color perception. Sample Data: The colors in the painted circles should be self-evident but will not always be the same from one student’s paper to the next; the final colors depend on how each student chose to mix the colors. Procedure 6: Any localized light source can be used as long as all light coming from it passes through the color filters provided. If the room is brightly lighted, you may again need a cardboard box to place over the light and the samples under study to prevent contamination by ambient white light. The procedure calls for the use of the colored circles that the students have just finished painting. This could be messy. You may wish to have a sample of this sheet prepainted to show as an example when explaining Procedure 5 to the students. This dry sheet may then be used by all groups in Procedure 6. Your dry sample sheet will be quite similar to those produced by the individual students, so the observed results will be much the same. Instruct the students to watch a particular colored circle as the filter is moved in front of the white light source. When the red filter is used, the reds and related colors should remain relatively unchanged, but the blues and greens will appear darker. (They would become black if the painted colors and the filters were perfect.) Similar results should be obtained with the blue and green light filters. The row of gray tones will show a general tinting toward the color of the light filter used. This shift occurs because all three primary colors are contained in gray, and so some reflection of each primary color can occur. Sample Data: As explained above, the color of an object depends on the colors of light available to reflect from its surface as well as on the colors absorbed by the surface itself. Some colored circles will change color, but others will remain nearly the same when each color filter is placed over the white light source.

ANSWERS TO PROCEDURE 5 IN THE LAB MANUAL 1.

Do there appear to be any differences in color between the two identical gray paint dots that you applied to the white paper and to the black paper? Explain. Ans: The same tone of gray does appear different on black paper than on white paper because background coloration affects color perception. The gray on white will appear darker in color, whereas the same gray applied on black will appear lighter in color.

ANSWERS TO PROCEDURE 6 IN THE LAB MANUAL 1.

(a)

(b)

Place a red-colored filter over a high-intensity lamp, and allow the light to fall on the piece of white paper on which you painted the dots in Procedure 5. Note any major changes in the colors of the dots. Remove and replace the filter several times while watching the color dots, and pick out the ones that show the most dramatic color changes and describe these changes below. (Note: Your instructor may provide you with a set of previously painted dots or other color wheel diagram to use because your paint may not be dry from your work in Procedure 5.) Why do you think changes are observed in the different shades of gray when the red filter is placed in front of the white light source? Ans: (a) For a red filter, reds and colors containing red remain nearly the same. Greens and blues turn darker. The extent of change depends on the quality of the color filter, but changes should be fairly consistent with these two statements. (b) Because white reflects all colors and red is the only color of light present, the different shades of gray take on a pinkish tinge.

Now place a blue filter over the light and record any changes in the colors of the dots. Ans: For a blue filter, blues remain nearly the same. Greens and reds turn darker.

Repeat the procedure with a green filter and record any color changes of the dots. Ans: For a green filter, greens remain nearly the same. Blues and reds turn darker.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Name the color hues found in the spectrum of white light. Ans: Red, orange, yellow, green, blue, and violet

Name the primary and secondary colors of the additive theory of light. Ans: Primary colors are red, green, and blue. Secondary colors are cyan, magenta, and yellow.

Name the primary and secondary colors of the subtractive theory of color. Ans: Primary colors are cyan, magenta, and yellow. Secondary colors are red, green, and blue.

If the primary colors of the subtractive theory are cyan, magenta, and yellow, why are the colors in the artist’s subtractive theory red, blue, and yellow? Ans: Pigments of cyan and magenta are not found naturally and are not easy to manufacture. However, red and blue pigments are readily available. This means that red and blue have been used traditionally by artists as their “primary” colors. A wider range of colors could be mixed by painters if cyan, magenta, and yellow were used, but the availability of pigment had led to the continued use of red, blue, and yellow as the primary colors in the artist’s subtractive theory. 61

How do colors differ when viewed under incandescent light and fluorescent light? Ans: Because incandescent light contains a higher ratio of reds and oranges, these colors predominate when seen under this type of light. Most objects have a “warm coloration” when seen under incandescent light. Fluorescent light often contains higher ratios of blue and green; a “cooler” appearance is observed because of the domination of these shorter wavelength colors. Modern “cool white” fluorescent lamps do not produce this effect as much as older or less expensive fluorescent lamps do. The newer tubes have been carefully color-corrected by adjusting the phosphors found on the inside of the tube to give off light that more closely approximates noontime sunlight and is thus closer to incandescent light.

Experiment 17

Static Electricity INTRODUCTION A brief lecture on the care and use of the aluminum-leaf or gold-leaf electroscope is a must for this experiment; otherwise, the electroscope may be damaged. Caution the students never to touch the knob of the electroscope with any charged body. Also, if the weather is damp, a drying agent will have to be placed in the electroscope’s metal and/or glass case to help retain a charge on the leaves. There will be sufficient time to show a film or perform a laboratory demonstration with this experiment. Faraday’s ice-pail experiment and the Van de Graaff electrostatic generator are good demonstrations for a small group of laboratory students. Another alternative is to include part of Experiment 18 with this experiment because Experiment 18 requires considerable time.

ANSWERS TO PROCEDURE 1 DATA TABLE 17.1 Answer A

Repelled

Answer B

Repelled

Answer C

Attracted

Answer D

Attracted

Answer E

Attracted

Answer F

Repelled

Answer G

Like charges repel Unlike charges attract

Procedure 2 is purely observational, no questions.

ANSWERS TO PROCEDURE 3 1.

Determine the charge on a glass rod after rubbing it with silk. The charge on the glass rod is _______________ (positive or negative). The charge on the silk cloth is _______________ . Ans: positive, negative

Rub each of two inflated toy balloons with wool or fur. Will they attract each other or repel each other? Explain why. Ans: They will repel each other. They have like charges.

Determine the charge on each balloon using the electroscope when it has been charged positively. (Hint: Do the leaves spread farther apart or collapse when the charged object is brought near the electroscope knob?) Ans: Both balloons will have a negative charge.

Determine the charge on the wool or fur after it has been used to rub a rubber rod. Ans: Positive

(a) (b) Ans:

Obtain a few (four or five) pieces of paper 1 or 2 cm in size. Remove any electric charge by touching them with your hands. Test for any charge on the pieces of paper by bringing them near the knob of the charged electroscope. The pieces of paper are neutral. That is, they have zero charge. Place them on the table free of any electric charge. Place a negative charge on a rubber rod by rubbing it with wool or fur. Bring the (a) charged rubber rod near the neutral pieces of paper. What do you observe? Explain. Repeat using a glass rod and a silk cloth. What do you observe? Explain. (b) Ans: (a) Attraction. The paper is charged in regions (polarized) by induction. Attraction takes place because the region closest to the charged rod is opposite (+) in sign. Attraction. Regions are polarized, as they were in Procedure 6 (a). But in the (b) reverse way so the two still attract.

Place the two metal spheres mounted on insulated stands in contact with one another as shown in Fig. 17.4, Part 1. A rubber rod carrying a negative charge is placed near the left side of sphere A. The spheres are then separated while the rubber rod is held near sphere A. See Fig. 17.4, Part 2 below. The spheres will now possess an electric charge if they are separated before the charged rod is removed. Determine what kind of charge will be on each by moving each in turn near the knob of a positively charged electroscope. Explain why each sphere was charged the way it was.

What kind of charge will a sheet of paper have after it has been rubbed with wool or fur? What kind of charge will a sheet of paper have after it has been rubbed with silk? (a) Negative Negative, electrons are transferred to the paper in both cases. (b)

Record the charge you detected on each sphere. Sphere A _______________________ Ans: Positive Sphere B________________________ Ans: Negative 64

Part 1

Part 2 Figure 17.4

Ans:

The two spheres will possess an electric charge. Sphere A will possess a positive charge, and sphere B will possess a negative charge, because the negative-charged rubber rod repels the negative charges on sphere A over to sphere B. This places an excess of electrons on sphere B and a deficiency on sphere A. The spheres are then separated, leaving them charged as indicated.

ANSWERS TO QUESTIONS 1.

What kind of a charge can be placed on an electroscope by induction using a glass rod rubbed with silk? Explain. Ans: A negative charge is placed on an electroscope by induction when a glass rod rubbed with a silk cloth is used to charge the rod. A glass rod, when rubbed with a silk cloth, will possess a positive charge. When the positive charged rod is placed near the knob of the electroscope, electrons will be attracted to the knob. While holding the charged rod near the knob, touch the knob with your finger. Electrons will flow from your body to the electroscope. Remove your finger from the electroscope. The electrons have no conducting path to leave the electroscope. Thus, the electroscope retains an excess of electrons and a negative charge.

How many different kinds of charges did you observe the effects of in this experiment? Ans: The effects of two different kinds of charges were observed in the experiment.

State the law of signs for electric charges. Ans: The law of signs states that like charges repel and unlike charges attract.

(a)

(b)

Ans:

Sometimes when a person combs their hair on a dry day (low relative humidity), a crackling sound can be heard as the plastic comb passes through the hair. What causes this to happen? Can you guess what type of charge (+ or –) will be left on the plastic comb after it has been passed through the hair? Will the hair be charged positively or negatively after the combing is complete? (a) Charge separation occurs as the comb passes through the hair. Subsequent passes of the comb bring the separated charges together again and the crackle is the electrostatic discharge as this recombination occurs. The comb should be negative (like the rubber rod rubbed with fur) and the (b) hair should be left with a positive charge.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Two charged objects attract one another with a force F. If the charge on both objects is doubled and the distance between them is also doubled, what will be the force between them? Ans: The force between them will be the same.

Two objects (for example, two toy balloons) in the laboratory are observed to attract each other. Can we be certain that both are electrostatically charged? Ans: No. Only one balloon must possess a charge. The other balloon could be charged by induction by the first charged balloon. It is also possible, however, that both balloons are charged, one positively and the other negatively.

Can we be certain that one of the balloons in Question 2 is electrostatically charged? Ans: Yes. At least one balloon must possess a charge in order to induce a charge on the other one, or as stated above both balloons could be charged.

Experiment 18

Magnetism and Electromagnetism INTRODUCTION Some introductory remarks about magnetism that include the following information are usually beneficial. Distinguish between the intensity of a magnetic field and the direction of a line of force. Indicate that a magnetic field is a region or volume of space surrounding a magnet within which the influence of the magnet is noticeable. The north-seeking end of a small compass placed in the region indicates the direction of the force field or the direction of a line of force at that point, and the forceper-unit pole is the intensity. In Procedure 2, point out to students that they are mapping the magnetic field resulting from Earth’s magnetism and the magnetism of the bar magnet. These two magnetic fields combine to produce the resultant field that the students are mapping using the small compass. Stress the point that magnetic lines of force are imaginary lines that show the direction of the resultant magnetic field. DATA TABLE 18.1 Procedures 1 and 2 Answer A: Repel Answer B: Attract Answer C: Attract Answer D: Repel Answer E:

Like poles repel; unlike poles attract.

Answer F:

The force that influences the suspended magnet becomes greater as the distance between them is decreased.

DATA TABLE 18.2 Procedure 3 Position A

Out of page, toward student

Position B

Into page, away from student

Position C

Out of page, toward student

Position D

Into page, away from student

Position E

Into page, away from student

Position F

Out of page, toward student

ANSWERS TO QUESTIONS IN THE MANUAL 1.

State the general rule for the interaction of magnetic poles. Ans: Like poles repel; unlike poles attract.

State Coulomb’s law for magnetic poles. Ans: The force of attraction or repulsion between magnetic poles is directly proportional to the product of their pole strength and inversely proportional to the distance between the poles.

How is the direction of a magnetic field defined? Ans: The direction of a magnetic field is the direction the north seeking pole of a compass tends to point when placed in the field.

What is the polarity (north or south) of Earth’s magnetic pole which is located near Earth’s north geographic pole? Explain. Ans: Near Earth’s north geographic pole is a south magnetic pole. A compass needle aligns its long axis with Earth’s magnetic field. The north-seeking pole is called the north pole of the magnet. Because unlike poles attract, the magnetic pole near Earth’s geographic pole therefore must be a south magnetic pole.

What is magnetic declination? Ans: Magnetic declination is the angle that a compass needle deviates from geographic north.

What is the angle of dip? Ans: The angle of dip is the angle Earth’s magnetic field makes with Earth’s surface (the horizontal) at a given latitude.

What is the magnitude of the angle of dip at Earth’s magnetic north pole? Ans: At Earth’s magnetic north pole the angle of dip is 90°.

Explain why the lines of force never cross one another. (Hint: Consider the direction of a magnetic line of force.) Ans: If lines of magnetic force crossed one another, the point of intersection would show two different directions of the magnetic field at the intersecting point. This is impossible.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Two iron bars are on the laboratory table. They are identical in every respect except that one is a magnet and the other is not. Without using anything other than the two iron bars, explain how you could identify the magnet. Ans: The nonmagnetized iron bar will neither attract nor repel the center section of the magnetized bar, but the magnetized iron bar will attract the center section of the nonmagnetized bar.

Do magnetic lines of force represent lines of force along the entire length of which the force on a unit pole would have the same value? Explain your answer. Ans: No. The force on a unit pole depends on the strength of the magnet producing the magnetic field and on the distance from the magnet.

Is it possible for two force lines of a magnetic field to intersect? Explain. Ans: No. If two lines intersected, it would indicate that the magnetic field pointed in two directions at the point of the intersection.

Experiment 19

Ohm’s Law INTRODUCTION Most students have trouble visualizing an electric circuit and have fears of getting shocked when working with electrical components and circuits. It is best to give an introduction to electricity and specifically to Ohm’s law before starting the experiment. This introduction is necessary if the students have not covered this material in the lecture class. Safety should be stressed in terms of both harm to students and damage to electrical components, especially the ammeters. Emphasize the difference between amount of charge (coulombs) and rate of flow of charge (amperes), since students often have difficulty with these two concepts. No data are provided for this experiment. The voltage and current values vary depending on the supply voltage and the maximum value of the milliammeter. Graphs 1 and 2, that are required for calculations 1 and 2 in the Laboratory Guide, will produce straight lines. The slope of each will yield the value of the resistor used in the circuit.

ANSWERS TO QUESTIONS IN THE MANUAL 1.

If you connect the circuit as shown in Fig. 19.1 in the Laboratory Guide with either resistor in place, but one fuse is removed from the circuit, this represents a circuit with a burned-out fuse. What is the voltage across the resistor? Explain your answer. V = ___________ Ans:

The voltage across the resistor is zero. Because no current is flowing through the resistor and one conducting path to the voltage source is open, the voltage across the resistor will be zero.

What is the voltage across the fuse that was left in the circuit? Explain your answer. V = ___________ Ans:

The voltage across the fuse in the other fuse holder will be zero. The explanation is the same as in Question 1. That is, no current is flowing through the fuse, and there is no conducting path to the other side of the voltage source.

What is the voltage across the terminals of the fuse holder without the fuse? Explain your answer. V = ___________ Ans:

The voltage across the fuse holder without a fuse will be nearly equal to the supply voltage. There is a conducting path from each terminal of the fuse holder back to the voltage source. Therefore, current will flow in the circuit when the voltmeter is connected. The internal resistance of the voltmeter is much higher than the resistance in the circuit shown; therefore, most of the supply voltage will appear across the voltmeter.

A voltmeter reads 80 volts when connected in parallel with an unknown resistor that has 125 milliamperes flowing through it. What is the resistance of the unknown resistance? Show your work. Rx = ___________

Ans:

Rx =

V 80 V = = 640Ω . A I 0125

Distinguish between an open and a closed circuit. Give an example of each. Ans: An open circuit is one in which a complete electronic path does not exist to allow electrons to flow. An example is a circuit with an open switch. A blown fuse and a broken wire are other examples. A closed circuit is one in which there is a complete path through which electrons can flow. An example is the complete path provided in a lighting circuit when the light switch is closed.

A 60-watt light bulb operates on regular house voltage of 120 volts. Normal current through the bulb is 0.5 amp. Determine the resistance of the bulb’s filament. Show your work. Rx = ___________

Ans:

Resistance =

voltage 120 volts = = 240 ohms current 0.5 amp

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

The amount of electrical charge passing through a resistor in 10 s is 16 C. What is the current flow? Q 16 C I= = = 16 . A Ans: t 10 s

The current flow in a resistor is 2 A. Determine the amount of electric charge that passes through the resistor in 4 s. Ans: Q = It = 2 A × 4 s = 8C

The relationship between wattage, voltage, and current is given by the following equation: Watts = voltage × current. Is the resistance of the filament of a 100-watt light bulb more or less than the resistance of the filament of a 60-watt light bulb? Calculate the resistance of the filament of a 60-watt light bulb that operates on 120 volts. How does your answer compare with the answer to Question 6? V W 60 watts R= where I = = = 0.5 amp Ans: I V 120 volts 120 volts R= = 240 ohms 0.5 amp This is the same value found in Question 6. For the 100 watt bulb the resistance turns out to be 144 ohms, so its resistance is less than that of the 60 watt bulb.

Experiment 20

Electric Circuits INTRODUCTION The laws for analyzing electric circuits were formulated in the nineteenth century by Gustav Kirchhoff (1824-1887), a German physicist. A circuit junction is a point where three or more conductors come together. A circuit loop is any closed path that returns to the same point. Kirchhoff’s laws are an expression of conservation laws applied to electric circuits. The laws state (1) the sum of the currents entering a junction equals the sum to the currents leaving the junction, and (2) the sum of the electromotive force around a circuit loop equals the sum of the potential differences across the resistances around the loop. These laws form the bases for this experiment. The following are the answers to all measurements and questions asked for in Procedures 1 through 6.

ANSWERS TO QUESTIONS IN THE LAB MANUAL PROCEDURE 1 (a)

Connect the milliammeter in series with a 47-ohm resistor R1 to the power source as shown in Fig. 20.3. If you do not have a 47-ohm resistor, use another with a similar value. R1 =

Figure 20.3

ohms

Series circuit with one resistor, a milliammeter, and connected to a power source. 75

(b)

Measure the total current flowing in the series circuit. IT = 67 mA = 0.067 amps

(c)

Measure the voltage across the power source. VB =

(d)

volts

Measure the voltage across the resistor R1. V1 =

(e)

3.2

3.1

volts

Calculate the resistance of R1 using Ohm’s law. R1 =

3.1 volts V1 = = 46 ohms I T 0.067 amps

How does the value of the voltage across the power source VB compare with the voltage across the resistor R1? Ans: They are very nearly the same.

Find the percentage error between the value printed on resistor R1 and the value calculated in Procedure 1(e). 47 ohms − 46 ohms 1 Ans: Percent error = × 100 = × 100 = 2.1% 47 ohms 47

PROCEDURE 2 (a)

Add a second 47-ohm resistor R2 in series with R1 and the milliammeter. If a 47-ohm resistor is not used, fill in a different value for R2. R2 = 47 ohms. See Figure 20.1.

(b)

Measure the total current. IT =

(c)

(d)

0.034

amps

Measure the voltage across each resistor. V1 =

1.6

volts

V2 =

1.5

volts

Measure the voltage across both resistors at once. Note: This is the same as the voltage supplied by the voltage source. V1+ 2 = VB =

mA =

3.1

volts

(e)

Calculate the resistances R1, R2, and R1 + 2. R1 =

16 . volts V1 = R1 = = 47 ohms 0.034 amps IT

R2 =

15 . volts V2 = R2 = = 44 ohms 0.034 amps IT

R1+ 2 =

(f)

3.1 volts VB = R1+ 2 = = 91 ohms 0.034 amps IT

Calculate the value of the series combination of resistors R1 and R2 with the series resistor equation using the values marked on the resistors. R1+ 2 = R1 + R2 . R1+ 2 = 47 ohms + 47 ohms = 94 ohms

How well do the values for the resistors R1 and R2 calculated in Procedure 2(e) compare with the value marked on them by the manufacturer? R(2), 6 percent error. Ans: R(1), 0 percent error;

Calculate the percentage difference between the value calculated for R1+ 2 in Procedure 2(e) and the value calculated for the series combination of these two resistors R1+ 2 found in Procedure 2(f). 94 ohms − 91 ohms × 100 = 3.2% Ans: Percent difference = 92.5 ohms

PROCEDURE 3 (a)

Add a third 47-ohm resistor R3 in series with R1, R2, and the milliammeter. If a 47-ohm resistor is not used, fill in a different value for R3, R3 = 47 ohms

(b)

Measure the total current. IT = 22 mA = 0.022 amps

(c)

Measure the voltage across each resistor. V1 = 1.1 volts V2 = 1.1 volts V3 = 1.1 volts

(d)

Measure the voltage across all three resistors in series at once, and record this value below. Note: This is also the value for the voltage supplied by the voltage source, VB. V1+ 2 + 3 = VB = 3.2 volts

(e)

Calculate the resistances R1, R2, R3, and R1 + 2 + 3. R1 =

11 . volts V1 = R1 = = 50 ohms 0.022 amps IT

R2 =

11 . volts V2 = R2 = = 50 ohms 0.022 amps IT

R3 =

11 . volts V3 = R3 = = 50 ohms 0.022 amps IT

R1+ 2 + 3 =

(f)

3.2 volts VB = R1+ 2 + 3 = = 145 ohms 0.022 amp IT

Calculate the value of the series combination of resistors R1, R2, and R3 with the series resistor equation using the values marked on the resistors. R1+ 2 + 3 = R1 + R2 + R3 R1+ 2 + 3 = 47 ohms + 47 ohms + 47 ohms = 141 ohms

Calculate the percentage difference between the values calculated for R1+ 2 +3 in Procedure 3(e) and the value calculated for the series combination of these three resistors R1+ 2 +3 found in Procedure 3(f). 145 ohms − 141 ohms Ans: Percent efficiency = × 100 = 2.8% 143 ohms

Add the voltages measured across the three resistors V1, V2, and V3 together and compare this sum with the value measured for V1+ 2 + 3 in Procedure 3(d). Ans: This calculation gives about three percent difference.

Calculate the IR drops across the three resistors in the series circuit and compare them with the voltages measured across each resistor. Ans:

I T R1 = 1.1 volts

V1 = 1.1 volts

I T R2 = 1.1 volts

V2 = 1.1 volts

I T R3 = 1.1 volts

V3 = 1.1 volts

Add the three values for the IR drops together and compare them with the value for the voltage across the entire series circuit. Ans: Sum of 3 IR drops = 3.3 volts VB = 3.2 volts

PROCEDURE 4 (a)

Replace two of the resistors in the series circuit used in Procedure 3 with a 10-ohm resistor and a 22-ohm resistor. This makes R1 = 10 ohms, R2 = 22 ohms, and R3 = 47 ohms. Again, if you do not have resistors of these exact values, use similar ones and fill in their values below. R1 = 10 ohms R2 = 22 ohms R3 = 47 ohms

(b)

Measure the total current. IT = 40 mA= 0.040 amps

(c)

Measure the voltage across each resistor. V1 = 0.40 volts V2 = 0.95 volts V3 = 1.95 volts

(d)

Measure the voltage across all three resistors at once. V1+ 2 + 3 = VB = 3.1 volts

(e)

Calculate the resistances R1, R2, R3, and R1+ 2 +3 . R1 =

V1 0.40 volts = = 10 ohms I T 0.040 amp

R2 =

0.95 volts V2 = = 24 ohms I T 0.040 amp

R3 =

. volts V3 195 = = 49 ohms I T 0.040 amp

R1+ 2 + 3 =

(f)

3.1 volts VR = = 78 ohms I T 0.040 amp

Calculate the value of the series combination of resistors R1, R2, and R3 with the series resistor equation using the values marked on the resistors. R1+ 2 + 3 = R1 + R2 + R3 R1+ 2 + 3 = 10 ohms + 22 ohms + 47 ohms = 79 ohms

Calculate the percentage difference between the values calculated for R1+ 2 +3 in Procedure 4(e) and the value calculated for the series combination of these two resistors R1+ 2 +3 found in Procedure 4(f). 79 ohms − 78 ohms × 100 = 13% . Ans: Percent efficiency = 78.5 ohms

Add the voltages measured across the three resistors V1, V2, and V3 together and compare this sum with the value measured for V1+ 2 + 3 in Procedure 3(d). . volts = 3.3 volts Ans: V1+ 2 + 3 = 0.40 volts + 0.93 volts + 195 . volts about 6% error compared = N B = 31

Calculate the IR drops across the three resistors in the series circuit and compare them with the voltages measured across each resistor. V1 = 0.40 volts Ans: ITR1 = 0.40 volts ITR2 = 0.96 volts

V2 = 0.95 volts

ITR3 = 1.96 volts

V3 = 1.95 volts

These compare within less than 1% error. 79

Add the three values for the IR drops together and compare them with the value for the voltage across the entire series circuit. VB = 3.1 volts Ans: Sum of 3 IR drops = 3.3 volts

In general, how do you think the sum of all of the IR drops in a series circuit compares with the total voltage across the entire circuit based on your results in Procedures 2, 3, and 4? Ans: Sum of the IR drops = total voltage across the circuit to within about 6%

Why didn’t you have to take separate measurements for the individual currents through resistors R1, R2, and R3 in Procedures 2, 3, and 4? Ans: In a series circuit the current is the same in each component.

Do you think that this experiment confirms the validity of the series resistor equation used in Procedures 2(f), 3(f), and 4(f)? Ans: Yes, these results are well within the errors inherent in the reading accuracy of the electrical meters used.

PROCEDURE 5 (a)

Connect three 47-ohm resistors in parallel with the milliammeter as shown in Fig. 20.4. If 47ohm resistors are not used, fill in a different value in the space below. R1 = R2 = R3 = 47 ohms

Figure 20.4 Parallel circuit shown as it might look with three resistors connected using clip leads (ends marked with x). Ammeter can be moved to points 1, 2, and 3 to measure current through the individual resistors. (b)

Measure the total current. IT = 191 mA = 0.191 amps

(c)

Measure the voltages across each resistor. V1 = 3.2 volts V2 = 3.2 volts V3 = 3.2 volts The voltage is the same across each resistor.

(d)

Measure the voltage across the power source. VB = 3.2 volts

(e)

Now carefully move the milliammeter from point T in the parallel circuit to point 1, and measure the current through resistor R1 and record this value below. Now repeat by moving the milliammeter to point 2 and to point 3. Keep all three parallel resistors connected while you make each measurement. If you happen to have four milliammeters available, you can set up the circuit with meters at T, 1, 2, and 3 all at once and take the current readings in that way, but you will get acceptable results if you use one meter and carefully move it to each location in turn. Now calculate the resistances R1, R2, R3, and R1+ 2 +3 . I1 = 64 mA = 0.064 amps

R1 =

V1 = 47 ohms I1

I2 = 65 mA = 0.065 amps

R2 =

V2 = 46 ohms I2

I3 = 63 mA = 0.063 amps

R3 =

V3 = 48 ohms I3

R1+ 2 + 3 =

(f)

3.2 volts VB = R1+ 2 + 3 = = 16.7 ohms . amp 0192 IT

Calculate the value of the parallel combination of resistors R1, R2, and R3 with the parallel resistor equation using the values marked on the resistors. 1 R1+2 + 3

1 1 1 + + R1 R2 R3

R1+ 2 + 3 = 15.7 ohms

How well do the values calculated for the resistors R1, R2, and R3 compare with the value marked on them by the manufacturer? Ans: They are within 1 ohm.

Calculate the percentage difference between the values calculated for R1+ 2 +3 in Procedure 5(e) and the value R1+ 2 +3 calculated for the parallel combination of these three resistors found in Procedure 5(f). 16.8 ohms − 15.7 ohms × 100 = 6.8% Ans: Percent difference = 16.25 ohms

Add the currents I1, I2, and I3 that you measured flowing through the three resistors in Procedure 5(e), and compare this sum with the value measured for IT in Procedure 5(b). Ans: I1 + I 2 + I 3 = 0192 . amps I T = 0191 . amps They agree within about 0.5%.

Calculate the IR drops across the three resistors in the series circuit and compare them with the voltages measured across the individual resistors V1, V2, and V3. V1 = 3.0 volts Ans: I1R1 = 3.00 volts I2R2 = 2.99 volts

V2 = 3.0 volts

I3R3 = 2.97 volts

V3 = 3.0 volts

Again these compare very well.

PROCEDURE 6 (a)

Replace one of the resistors in the parallel circuit used in Procedure 5 with an 82-ohm resistor and take the third one out of the circuit completely. This makes R1 = 47 ohms and R2 = 82 ohms. Again, if you do not have resistors of these exact values, use similar ones. R1 = 47 ohms

(b)

R2 = 82 ohms

Measure the current flowing through both resistors at once. IT = 100 mA = 0.10 amps

(c)

Measure the voltage across each resistor. V1 = 3.0 volts V2 = 3.1 volts

(d)

Measure the voltage across the voltage source. VB = 3.1 volts

(e)

Measure the current flowing through each resistor I1 and I2. You may have to move the milliammeter around in the circuit if you only have one meter. Then calculate the resistances R1, R2, and R1 + 2. I1 = 66 mA = 0.066 amps

R1 =

V1 = 45 ohms I1

12 = 36 mA = 0.036 amps

R2 =

V2 = 86 ohms I2

R1+ 2 =

(f)

Calculate the value of the parallel combination of resistors R1 and R2 with the parallel resistor equation using the values marked on the resistors. 1 R1+2

R1+ 2 =

. volts 31 VB = R1+2 = = 30 ohms 0.102 amps IT

1 1 1 1 1 + = = + = 0.022 + 0.012 = 0.034 R1 R2 R1+ 2 45 86 1 = 29 ohms 0.034

Calculate the percentage difference between the values calculated for R1+ 2 in Procedure 6(e) and the value calculated for the parallel combination of these two resistors R1+ 2 found in Procedure 6(f). 30 ohms − 29 ohms × 100 = 3.4% Ans: Percent difference = 29.5 ohms

Add the currents you measured flowing through the two resistors I1, and I2 together and compare this sum with the value measured for IT in Procedure 6(b). IT = 100 mA. Ans: I1 + I2 = 66 mA + 36 mA = 102 mA; Again nearly the same.

Calculate the IR drops across the two resistors in the parallel circuit and compare them with the voltages measured across each resistor V1 and V2. V1 = 3.0 volts Ans: I1R1 = 3.0 volts I2R2 = 3.1 volts

V2 = 3.1 volts

The calculated values are the same as the measured values. 4.

After reviewing the data from Procedures 5 and 6, in general how do you think the sum of all the currents flowing through the resistors in a parallel circuit compares with the total current flowing in the circuit? Ans: The sum of all currents = I T

After reviewing the data from Procedures 5 and 6, in general how do you think the voltages across individual elements in a parallel circuit compare with the total voltage supplied to the circuit? Ans: The supply voltage = voltage across each component.

Why do you have to take separate measurements for the individual currents through resistors in Procedures 5 and 6? Ans: The currents are not equal. The current through each component depends on the resistance of each component.

Does this experiment confirm the validity of the parallel resistor equation used in Procedures 5(f) and 6(f)? Ans: Yes, quite well.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Are the light bulbs in your home, say in three separate lamps, connected in series or parallel across the 110 volt power line coming from the fuse box in your home? Ans: They are connected in parallel.

If 1.2 amps flows in each light bulb, what is the total current flowing in the circuit? I T = I1 + I 2 + I 3 = 3.6 amps Ans:

How much voltage appears across each bulb? Ans: Since they are in parallel, the full 110 volts appears across each bulb.

What is the internal resistance of each light bulb? V 110 volts Ans: R1 = 1 = = 92 ohms I1 12 . amps

Experiment 21

Electromagnetic Waves INTRODUCTION Although the term electromagnetic waves may not be familiar to all students, most are certainly acquainted with the terms heat, light, microwaves, radio, TV, and X-rays. Since these terms are common in our daily life, this experiment’s objective is to bridge the gap between informal experience and scientific understanding. We cannot expect students taking this physical science course to understand Maxwell’s electromagnetic field theory, but we can stress the fact that electromagnetic waves are generated and that energy is propagated from accelerated charges. Emphasize this fact at every opportunity you have with your students. Note that column two in Table 21.2 repeats this fact for each type of radiation. This is strictly a math-oriented experiment. Where possible, it is suggested that you display an LC circuit (an old uncased portable radio) or other devices for generating and detecting electromagnetic waves. Display an ultraviolet lamp, a Geiger tube, an old X-ray tube, or any of the devices listed in column two and three in Table 21.2 if you have them available. Many students have trouble plotting graphs. One of the major problems is the choice of scale for the x and y axes. The graph in this experiment deals with very large numbers on the y axis and very small numbers on the x axis, so students may need some assistance from the instructor on how to select the scales for the two axes.

ANSWERS TO PROCEDURE 1

Frequency f (Hz)

DATA TABLE 21.1 Calculated Wavelengths for Visible Light Frequencies Wavelength λ Wavelength λ Wavelength λ (m) (Å) (nm)

Color

4.3 × 1014

8 × 10–7

8000

800

Red

5.0 × 1014

6 × 10–7

6000

600

Orange

6.0 × 1014

5 × 10–7

5000

500

Green

7.5 × 1014

4 × 10–7

4000

400

Violet

ANSWERS TO PROCEDURE 2 1.

State in words the relationship between frequency and wavelength that your graph illustrates. Ans: This is an inverse relationship; as the frequency increases, the wavelength decreases.

Refer to Experiment 1 in this Laboratory Guide and give the name that describes the curve in your graph. (Hint: What is the equation for the curve?) Ans: Hyperbola

ANSWERS TO PROCEDURE 3

Name of Radiation Radio waves AM broadcast band FM broadcast band TV broadcast band Microwaves Infrared (heat waves)

DATA TABLE 21.2 Electromagnetic Radiations Wavelength Range Meters Frequency Range to be Calculated Hertz (cycles/s) by Student Approximate Values 0.55 × 106 to 160 . × 106 88 × 106 to 108 × 106 54 × 106 to 890 × 106 1 × 109 to 1 × 1011

545 3.40 5.55 3 × 10–1

to to to to

188 2.78 0.34 3 × 10–

1 × 1011 to 4.3 × 1014

3 × 10–3

8 × 10–

Visible light

4.3 × 1014 to 7.5 × 1014

8 × 10–7

4 × 10–

Ultraviolet

7.5 × 1014 to 3 × 1017

4 × 10–7

1 × 10–

X-rays

3 × 1017 to 3 × 1019

1 × 10–9

1 × 10–

Gamma rays

Greater than 3 × 1019

Less than 10–11

AM FM TV MW

ANSWERS TO PROCEDURE 4 Calculate the energy of a photon in Joules with an associated frequency listed below: (a)

2 × 1014 Hz E = hf = 6.63 × 10 −34 J-s × 2.0 × 1014 cycles s = 13 . × 10 −19 J (infrared) Ans:

(b)

4.3 × 1014 Hz E = hf = 6.63 × 10 −34 J-s × 4.3 × 1014 cycles s = 2.9 × 10 −19 J (red light) Ans:

(c)

7.5 × 1014 Hz E = hf = 6.63 × 10 −34 J-s × 7.5 × 1014 cycles s = 5.0 × 10 −19 J (violet light) Ans:

(d)

3 × 1018 Hz E = hf = 6.63 × 10 −34 J-s × 3 × 1018 cycles s = 2 × 10 −15 J (X-rays) Ans:

ANSWERS TO QUESTIONS 1.

What is the basic explanation for the origin of electromagnetic waves? Ans: The acceleration of an electrically charged particle (electron, proton, or ion) will release energy that spreads outward as electromagnetic waves.

All electromagnetic waves have the same physical characteristics. They differ only in frequency and wavelength. How are they classified into different types? Ans: Basic names are given to electromagnetic waves depending on the frequency and wavelength ranges within which they fall. They are further classified by the method by which they are generated or detected.

How do you account for the fact that a sharp dividing line does not exist between one type of electromagnetic wave and its neighbor? Ans: The methods of generating the radiation overlap one another.

How are radio waves generated? How are they detected? What do you think limits the highest radio frequency that can be generated? Ans: Radio waves are generated and detected by an LC circuit. The highest frequency is limited by the physical dimensions of the LC circuit.

Why is it impossible to generate radio waves with a frequency of 4 × 1014 Hz ? Ans: An LC circuit cannot be made small enough for this frequency.

Use the data given in Table 21.2, Fig. 21.1, and Fig. 21.2 to answer Questions 6, 7, and 8. 6.

How are X-rays generated? Give three methods for detecting them. Ans: X-rays are generated by disturbed electrons in atoms. They are detected by photographic film, Geiger tubes, ionization chambers, and fluorescent screens.

What is the minimum frequency generated by radioactive nuclei? The maximum wavelength? What is the name of this radiation? f = 3 × 1019 Hz , λ = 1 × 10 −11 m , gamma rays Ans:

Name the type of radiation that has the greatest range of frequencies. Give the range width. Ans: Infrared; 4.3 × 1014 Hz = 4300 × 1011 Hz ; 4300 × 1011 Hz − 1 × 1011 Hz = 4299 × 1011 Hz of frequency range width.

Electromagnetic waves are transverse waves. State the relationship between the electric, magnetic, and wave velocity vectors. Ans: The three vectors are 90° to one another.

10.

Calculate the difference in energy (joules) between red and violet light. Ans: Refer to the answers to Procedure 4. 5.0 × 10 −19 J − 2.9 × 10−19 J = 2.1 × 10 −19 J

11.

Does a photon associated with blue light have more or less energy than a photon associated with red light? Give the reason for your answer. Ans: A photon of blue light has more energy because its frequency is higher so E = h f means more energy per photon.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

The electrical energy supplied to this laboratory is specified as 120 volts ac, 60 Hz. Calculate the wavelength associated with this frequency. c 3 × 108 m s Ans: Wavelength = = = 5 × 106 m f 60 Hz

What is the phase relationship between the electric vector and the magnetic vector of electromagnetic waves? Ans: The two fields are exactly in phase.

Give the SI units for velocity, wavelength, and frequency. Ans: Velocity is in m/s, wavelength is in m, frequency in cycles/second or Hertz

How long (time in seconds) does it take light (electromagnetic radiation) to travel from the Sun to Earth? See your textbook (inside back cover) for the distance between the Sun and Earth. d 15 . × 1010 m = 5 × 102 s or about 8 minutes Ans: Time = = t 3 × 108 m s

The term radar is an acronym for radio detecting and ranging. Radar generators produce and emit short bursts of electromagnetic waves, which are reflected by objects and detected by the radar receiver. Determine the wavelength of a radar frequency of 5.3 GHz. (a) Refer to Data Table 21.1 and give the name of this radiation. (b) 3 × 108 m s = = 5.7 × 10 −2 m Ans: (a) Wavelength 9 5.3 × 10 Hz (b) Radio (microwaves)

The term laser is an acronym for light amplification by stimulated emission of radiation. If a certain laser generates a wavelength of 1.3 µm. (a) Determine the frequency associated with this wavelength. (b) Refer to Data Table 21.1 and give the name of this radiation. 3 × 108 m s Ans: (a) Frequency = = 2.3 × 1014 Hz −6 13 . × 10 m (b) Infrared

Experiment 22

Temperature INTRODUCTION In this experiment students can discover how the Celsius temperature scale was devised and constructed by calibrating an alcohol-in-glass or a mercury-in-glass thermometer. The experiment can be completed in 1 hour or less by most students. If a film on heat or temperature is not available, Experiment 21 or 22 also could be done during this laboratory period using the thermometer calibrated in this experiment. One can present temperature as the concept having a great influence on students’ daily lives. Air temperature influences our mood, dress, homes, and activity. No data are supplied for this experiment since nearly all of this experiment is observational and revolves around he construction of a simple thermometer.

ANSWERS TO QUESTIONS IN THE MANUAL 1.

What is the least count of your hand calibrated thermometer? Ans: One degree Celsius.

What is the percent error of your newly calibrated thermometer in your determination of the existing air temperature in the laboratory? Use the reading on the standard commercially calibrated thermometer as the accepted value. Show your work. Percent error = ____________ Ans:

The percentage error is equal to the absolute difference between the temperature measured with the laboratory thermometer and the student’s calibrated thermometer divided by the laboratory thermometer reading × 100.

Would you expect the steam point on your calibrated thermometer to be higher or lower than a regular standard thermometer? Why? Ans: The steam point determined by the student’s thermometer may be lower because the standard thermometer is calibrated at sea level where the air pressure is greater.

What are the disadvantages of using a water-in-glass thermometer to measure outside air temperature? Ans: The major disadvantage is that water exists only between 0° and 100°C so it might freeze or boil in the thermometer. 89

When a mercury-in-glass thermometer is placed in boiling water, the mercury level falls slightly at first and then begins to rise. Explain this effect. If you have lab time, perform the experiment. Obtain the lab instructor’s permission. Ans: Because the mercury is enclosed in the glass bulb, the glass bulb is the first to become hot and expand. The larger glass volume allows the mercury to fall. Once the mercury itself begins to receive heat, it expands rapidly and rises again to the maximum level.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Explain the meaning of temperature. Ans: Temperature is a measure of how hot or cold an object is, or more specifically, temperature is proportional to the average kinetic energy of the molecules that make up the object.

The relationship between the Celsius and the Fahrenheit temperature scales is given by the following equation: C = 5 9 ( F° − 32° ) . Normal body temperature is 98.6°F. Determine the equivalent Celsius temperature. Show your work. C = 5 9 ( 98.6°F − 32° ) = 37°C Ans:

Explain what causes the mercury in the glass thermometer to rise and fall with a change in temperature. Ans: The mercury expands or contracts in volume when heat energy is added to or taken from the mercury and this causes its heat to vary in the thermometer.

At what temperature on the Fahrenheit temperature scale does water at atmosphere pressure boil? Ans: Water boils at 212°F.

Experiment 23

Specific Heat INTRODUCTION Briefly remark on the method of mixtures, emphasizing that there is no direct method for measuring heat. Although the term calorimeter means the measurement of heat, the device does not measure anything. Explain the difference between the calorimeter cup and the housing used to hold and insulate the cup from the atmosphere. Because the most critical part of the experiment is obtaining the final temperature of the mixture, emphasize number 5 in the Procedure in your remarks. Although better results for the specific heat of a metal can be obtained with metal pellets, a small solid object can be used. A percentage error of 5 percent can be obtained with careful taking of data. Students should make a second trial if the data do not yield results within 5 percent. If a regular calorimeter is not available, a coffee-cup calorimeter can be constructed using two polystyrene coffee cups placed one inside the other and held in a glass beaker. The top cup can be fitted with a Styrofoam lid with two cut holes for the thermometer and the stirrer. The following data are typical for the metal copper, which has a specific heat of 0.093 cal g °C .

DATA TABLE 23.1 Metal used: Copper Mass of the calorimeter cup and stirrer (1)

mcn

48 g

Mass of the calorimeter cup, stirrer, and water (2)

201 g

Mass of the calorimeter cup, stirrer, water, and metal (3)

273 g

Mass of the water [(2) − (1)]

mcw

153 g

Mass of the metal [(3) − (2)]

mmt

72 g

Temperature of the hot metal

Tmt

99° C

Temperature of the cold water

Tcw

22° C

Temperature of final mixture

25° C

ANSWERS TO QUESTIONS IN THE MANUAL 1.

Distinguish between heat and temperature. Compare definitions and units of measurements. Ans: Heat is energy in transit and is measured in calories or British thermal units. Temperature is a measurement of the hotness or coldness of an object, and temperature is proportional to the average kinetic energy of the molecules. Temperature is measured in degrees Celsius or degrees Fahrenheit.

State the specific heat of water in terms of standard SI energy units (joules). Ans: In the cgs system of units, specific heat is the amount of heat required to raise one gram of a substance one degree Celsius. Heat capacity is the amount of heat required to raise the temperature of the given mass (amount varies) one degree Celsius. Heat capacity = mass × specific heat.

The heat capacity of a material is defined as: heat capacity = mass of the material times its specific heat. What is the heat capacity in calories for: (a) 100 g of iron? (b) 2,500 g of water? Hc = mc = 100 g 0105 . cal g° C = 10.5 cal g° C Ans: (a) Hc = mc = 2,500 g 100 . cal g° C = 2500 cal g° C (b)

g g

If the final temperature was determined incorrectly (a value greater than the true value was obtained), how would this affect the calculated value of the specific heat? Explain your answer. Ans: The value for the specific heat will be much greater because the temperature difference for the water and the calorimeter cup will increase and the temperature difference for the metal will decrease. The specific heat for the metal is determined by dividing water and cup values by the metal values.

Where, do you feel, is the greatest possible source of error in this experiment? Why? Ans: The greatest possible source of error is the measurement of the final temperature of the mixture because the process of mixing the two substances evenly is difficult to accomplish.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

The specific heat of soil (average value) is one-fourth of a calorie per gram degree Celsius. Explain why water is a better moderator of temperature than soil. Ans: A higher specific heat means it takes more heat (and therefore more time) to raise the temperature of water by one degree. For example, the Sun will heat soil four times faster than it will heat water. When the heat source is removed, the soil will cool four times faster than the water.

Which has the greater specific heat, a pint of water or a gallon of water? The greater heat capacity? Explain. Ans: Both have the same value of specific heat. The gallon of water has the greater heat capacity. Heat capacity equals mass × specific heat. One gallon of water has more mass than one pint of water.

A metal having a mass of 65 g and a temperature of 100°C is placed in 150 g of water held in a coffee-cup calorimeter at 22° C. The highest temperature of the mixture was 25°C. (a) Calculate the change in the water temperature. (b) Calculate the change in temperature for the metal. Calculate the specific heat of the metal. (c) Ans: (a) ∆T for the water = 25°C – 22°C = 3° C (b) ∆T for the metal = 100°C – 25°C = 75°C 150 g × ( 25° − 22° ) × cw 450 × cw (c) Specific heat = = = 0.092 cal g °C 65 g × 100° − 25° 4875 cw = the specific heat of water = 1 cal g °C

What is specific about specific heat? Ans: A specific amount of mass. The standard one gram of any substance is the mass taken to determine the magnitude of the specific heat of that material in calories.

Experiment 24

Heat of Fusion INTRODUCTION The law of conservation of energy is used here to find an experimental value for the heat of fusion of ice. The method of mixtures is used to obtain the experimental value. A problem in this experiment will be random errors, and students will need special skills to keep them to a minimum. The greatest possible source of errors will be wet ice. Caution students that it is important for good results to have ice as free of water as possible. The experimental value for the heat of fusion should be within 5 percent of the accepted value of 80 cal/g. Require a second trial of a student who fails to obtain results within this limit. An aluminum calorimeter cup was used to obtain the information for Data Table 24.1. DATA TABLE 24.1 Mass of calorimeter cup and stirrer (1)

mcup

Mass of calorimeter cup, stirrer, and water (2) Mass of water [(2) − (1)]

48 g 248 g

Mass of cup, stirrer, water, and melted ice (3)

200 g 268 g

Mass of ice [(3) − (2)]

20 g

Initial temperature of water, cup, and stirrer

22° C

Final temperature of water, cup, and stirrer

13.5° C

∆T Tinitial − Tfinal

8.5° C

Specific heat of water

1.00 cal g°C

Specific heat of calorimeter cup and stirrer (obtain from instructor)

ccup

0.22 cal g°C

Heat of fusion of ice (experimental value)

81 cal g

CALCULATIONS FOR STEP 9 OF THE PROCEDURE Hf =

200 g× 1 cal g ⋅ °C ( 22°C − 13.5°C ) + 48 g × 0.22 cal g°C ( 22°C − 13.5°C ) 20 g

−20 g × cal g ⋅ °C ( 22°C − 13.5°C ) 20 g Hf =

( 200 × 8.5) cal + (10.6 × 8.5) cal − ( 20 × 8.5 ) cal

20 g 1700 cal + 90 cal − 170 cal 1620 cal Hf = = = 81 cal g 20 g 20 g

ANSWERS TO QUESTIONS IN THE MANUAL 1.

The accepted value for the heat of fusion of ice is 80 cal/g. This is the amount of energy that must be transferred into one gram of ice at its melting temperature (0°C) in order to change it into liquid water. Also, 80 cal/g is the amount of energy that must be removed from one gram of water at 0°C in order to change liquid water to ice. Calculate the percent error for your experimental value of the heat of fusion of ice using this value as the standard. Show your work. Ans: The percent error should be 5 percent or less. Refer to Appendix III for an equation to calculate percent error.

How will using wet ice (some water at 0°C is added along with the ice) affect the experimental value for the latent heat of fusion of ice? Explain. Ans: The calculated value for the heat of fusion of ice will be lower, because the mass of melted ice water in the experiment will be greater. See Eq. 24.3.

If the calorimeter is not well insulated, how will heat from the air in the laboratory affect the experimental value for the latent heat of fusion of ice? Explain. Ans: The calculated value for the heat of fusion of ice will be less because part of the heat transferred to melt the ice and raise the temperature of the melted ice water will be supplied from the air in the laboratory; therefore, not as much heat will come from the water and calorimeter cup. This will make the calculated value low.

How do you think citrus growers in Florida might use heat of fusion to protect against frost damage? Ans: Water sprayed on the citrus fruit requires that heat be removed from the water to change phase to ice. Eighty calories of heat must be removed per gram. This heat transfer warms to air temperature and this helps to keep the crop from freezing.

Is it possible to add heat energy to a substance and not increase its temperature? Give an example. Ans: Yes. The melting of ice.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Which has the greater effect in cooling a substance, 1 g ice at 0°C or 1 g water at 0°C? Explain. Ans: One gram of ice, because it will take 80 cal of heat to change phase from ice to water at 0°C.

Where are the greatest sources of error in this experiment? Ans: The greatest sources of error are keeping the ice dry and determining the final temperature accurately.

During the phase change of ice to water, the temperature remains constant. If energy cannot be destroyed, where does the supplied heat energy go? Ans: The supplied energy goes to separate the crystalline structure of the ice.

Is heat absorbed or released by ice when it melts? Ans: Heat is absorbed.

In what direction did the heat flow in this experiment? Ans: Heat flowed from the water (hot) to the ice (cold).

What basic thermodynamic law applies to Question 5? Ans: The second law of thermodynamics

Experiment 25

Heat of Vaporization of Water INTRODUCTION The method of mixtures is used in this experiment, and the law of conservation of energy applies. The steam gives up a certain amount of heat in condensing to water at its boiling point without a change in temperature. Caution the students to have a heavy flow of steam from the outlet tube and to lower the outlet to the bottom of the calorimeter cup. The steam flow causes turbulence in the water; therefore, no stirring is required. When the water temperature rises to about 40°C, remove the steam outlet from the water. Do not shut off the steam until the steam outlet has been removed from the water in the calorimeter cup. Also remind the students that the thermometer reading the steam temperature may not reach 100°C because the laboratory may have an elevation above sea level. A second trial is recommended if the results are not within 10 percent of the 540 cal/g. The following data for Data Table 25.1 are typical for the experiment.

DATA TABLE 25.1 Mass of calorimeter cup and stirrer (1)

mcup

Mass of calorimeter cup, stirrer, and water (2)

48.0g 248.0 g

Mass of water[(2 – 1)]

Mass of cup, stirrer, water, and condensed steam (3)

200.0 g 204.6 g

Mass of steam [(3) – (2)]

4.6 g

Initial temperature of water, cup, and stirrer

22° C

Final temperature of water, cup, and stirrer

36° C

∆T Tf − Tw

14° C

Specific heat of water

Specific heat of calorimeter cup and stirrer (obtain from instructor) Temperature of steam *If you are not given another value, use 99.0°C

ccup

Heat of vaporization of water (experimental value)

1.00 cal g⋅° C 0.22 cal g⋅° C 100° C 556 cal/g

Calculate the value for the heat of vaporization of water using Eq. 25.3. The final temperature of the condensed steam will be the same as the final temperature of the water in the calorimeter cup.

( mc∆T )water + ( mc∆T )cup = ( mH v ) + ( mc∆T )condensed steam 200.0 g (1 cal g ⋅ °C )(14°C ) + 48.0 g ( 0.22 cal g ⋅ °C )(14°C ) = 4.6 g H v + 4.6 g (1 cal g °C )(100°C ) = 36°C Hv =

100

2800 cal + 51 cal − 294 cal = 556 cal g 4.6 g

ANSWERS TO QUESTIONS IN THE MANUAL 1.

The accepted value for the latent heat of vaporization is 540 cal/g. This is the amount of energy that must be removed from one gram of steam at its boiling temperature (100°C) in order to change it into liquid water. Also, 540 cal/g is the amount of energy that must be added to one gram of water at 100°C in order to change liquid water to steam. Calculate the percentage error in your experimental value of Hv , using this value as the standard. Show your work. 556 cal g − 540 cal g Ans: Percentage error = × 100 = 3% 540 cal g

How will the experimental value be affected if some water (liquid) comes through the hose into the calorimeter cup along with the steam? Ans: The experimental value will be lowered.

If the calorimeter is not well insulated, how will heat from the air in the laboratory affect the experimental value for the latent heat of vaporization? Ans: A heat loss from the calorimeter cup will take place. Part of the heat energy from the steam will be lost, and the value for the heat of vaporization calculated will be lowered.

How can a steam turbine use the heat of vaporization to increase the output of electrical energy it is generating? Ans: The change in phase supplies 540 cal/g of heat energy to the process.

Is it possible to remove heat energy from a substance and not change its temperature? Explain, and give an example. Ans: Yes. The change in phase from a gas to a liquid, as shown by the condensation of steam in this experiment.

Calculate the percentage increase in the mass of the water in the calorimeter cup with the addition of the condensed steam. Use mass data from Table 25.1 for your calculation. Ans: Using sample data, the percent increase in the mass of water is ms mw × 100 = 2.3%

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What are some of the sources of error in the experiment? Failure to obtain the correct temperatures. Ans: (a) (b) Failure to determine the masses correctly. (c) Failure to have a good flow of steam. (d) Allowing condensed steam (water) from the supply system to enter the calorimeter cup.

101

102

What is the advantage of the steam trap? Ans: The steam trap collects condensed steam (hot water) and prevents most of it from entering the calorimeter cup.

Experiment 26

Radiation INTRODUCTION Because the introduction to this experiment is rather brief, give a few remarks on radioactive sources and their emitted radiation, along with an introduction to radioactive decay and half-life.

PROCEDURE 1.

The background count should be quite low but it is interesting to point out to the students that we live in a world where our bodies are continually exposed to external radiation. If the count rate for the sources used in this experiment is also quite low, the background count must be subtracted before the intensity is recorded in the tables. If it is only 1 or 2 percent, it can be ignored but proper procedure would be to subtract it anyway.

Count rate should decrease rapidly as the sources are moved away from the detector.

The thicknesses of the 3 shields should be about the same to show this effect properly. Then the counts/min should decrease for the beta particles for paper and wood but the gamma should remain about the same. The lead should shut off the beta particles entirely and reduce the gamma depending on its thickness.

A strong magnetic field should deflect beta particles (electrons) because of their negative charge and the count rate should reduce when the magnet is first in place. Gamma rays are not effected so no change should be observed.

The penetrating power of the radiation should allow you to distinguish easily between the two sources.

ANSWERS TO QUESTIONS IN THE MANUAL 1.

Why aren’t the three readings for background radiation in Procedure 1 identical? Ans: Background radiation comes from cosmic rays and local radioactive sources in a random way, and a time of 1 minute is too brief to obtain identical readings but these variations will average out over longer times.

103

Does your data for Procedure 2 suggest that there is a relation between 1 d 2 and the intensity? If so, what? Should there be a relationship? Discuss sources of error in this part of the experiment. Ans: A graph plotting 1/d2 versus intensity produces a straight line so there is a direct relationship. Possible sources of error: (1) failure to take enough time to determine an accurate radiation count, (2) inaccurate measurement of the distance between the radiation source and the center of the sensitive region of the Geiger tube.

Why are alpha particles so easily stopped by even very thin sheets of insulation material? Ans: Alpha particles are easily stopped because of their large size and because they carry a +2 positive charge.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Name the three distinct types of radiation emitted by radioactive sources. Ans: Alpha particles, beta particles, and gamma rays

State the physical characteristics of the three types of radiation from radioactive sources. Ans: Alpha particles are the nucleus of a helium atom and carry a +2 charge. Beta particles are electrons and carry a –1 charge. Gamma rays are electromagnetic radiation of very high frequency and carry no charge at all.

How could you test an ore sample for radioactivity? Ans: Use the Geiger tube and detector equipment used in this experiment.

How could you identify alpha, beta, and gamma rays coming from an ore sample? Ans: Pass the radiation from the ore sample through a magnetic field. The alpha and beta particles will be deflected in different directions. The magnetic field has no effect on the gamma rays or more practically use shielding material to make the tests like you did in Procedure 3.

104

Experiment 27

Spectroscopy INTRODUCTION As an introduction to spectroscopy and to this experiment, set up and demonstrate two student spectrometers. One spectrometer should have a grating with a large number of lines per centimeter, and the other, a smaller number to illustrate the type of spectrum produced by different values of the grating space. The demonstration also shows students that photographs of line spectra are photos of the slit illuminated by the source. Allow students to vary slit width while observing the spectrum lines, if possible. Take special care to shield the students’ vision as much as possible from the ultraviolet radiation coming from the mercury lamp. The students should be warned not to look directly at the mercury lamp. The following data for Table 27.2 in the Laboratory Guide were obtained using a sodium lamp and a grating with 7500 lines per inch. The grating space d = 1 7500 lines per in = 133 . × 10 −4 in or 3.38 × 10 −4 cm . DATA TABLE 27.2 Average of x1 x2 x2 Left and Left Right x1 Right (cm) (cm)

Average of x2 Left and x2 Right

Distance, s Between Grating and Slit (cm)

x1 Left (cm)

x1 Right (cm)

7.0

7.1

7.05

14.5

0.1750 0.3625

10.6

10.60

20.5

0.1770 0.3417

14.0

14.00

27.4

27.5

0.1750 0.3437

100

17.7

17.70

35.4

35.5

0.1770 0.3540

Average values of sin θ

∗

sin θ 1

∗

sin θ 2

0.1760 0.3505

105

ANSWERS TO CALCULATIONS 1.

Using nλ = d sin θ For the data for 80 cm in Data Table 27.2 −6 d sin θ 1 3.38 × 10 m 0.1750 λ= = = 5.915 × 10 −7 m n 1 1 avg = 591 avg 5.915 × 10 −7 m 1 × 10 −10 m Compared to the acceptable value, 5893 avg % error = 0.4% −6 d sin θ 2 3.38 × 10 m 0.3437 λ= = = 5809 × 10 −7 m . n 1 1 avg × 10 −7 m = 5809 avg 5809 . 1 × 10 −10 m with % error = 1.4%

FG H

b IJ K

FG H

b IJ K

. × 10−7 m Again using the acceptable value, 5893 avg = 5893

∆E =

6.67 × 10 −34 Js 3.00 × 108 m s 5893 . × 107 m

∆ E = 3.40 × 10 −19 J

ANSWERS TO QUESTIONS IN THE MANUAL 1.

What is a diffraction grating? Ans: A diffraction grating consists of many very fine, closely spaced parallel slits. The number of slits varies from a few hundred per inch to several thousand per centimeter.

What is the grating spacing? Ans: The grating space is the width of one slit.

In the first-order spectrum of mercury, if available for you to observe, is the red line or the violet line closest to the central image? Ans: The red line is closest to the central image. sin θ =

nλ d

and red light has the shorter wavelength (λ) which places it closer to the central image.

106

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Are the higher-order spectra more accurate than lower-order ones in determining wavelengths of light? Why or why not? Ans: The higher orders are more accurate because there is a greater spread of the spectrum lines and the measurements are generally more accurate.

One specific wavelength of red light is incident normally on a diffraction grating having 5000 lines/cm. If the first-order spectrum makes an angle of 22° with the normal of the grating, what is the wavelength of the red light? nλ = d sin θ Ans: 1 d= = 0.002 cm = 2 × 10−6 m 5000 line cm

λ = 2 × 10

−6

m × 0.375 1 = 0.7492 × 10

−6

!   ! 1A   = 7492 A m − 10  1 × 10 m   

What will be angle θ in Question 2 for the second-order spectrum? nλ 2 × 0.7492 × 10 −6 m sin θ = = = 0.749 Ans: d 2 × 10 −6 m θ = 48.5°

107

Experiment 28

Density of Liquids and Solids INTRODUCTION Density is a concept that is important in the science of chemistry and physics, and this experiment can be performed by the student when studying either branch of science. Note that the maximum density of pure air-free water occurs at 3.98°C. This is the temperature at which the density of water is . g cm3 . defined to be exactly 10000 Pycnometers are rather fragile, and students should be cautioned to handle them with care. Also advise students to remove all excess water from the surface of the volumetric flask before weighing. Because the data for this experiment vary greatly depending on the type and amount of sample used, no specific information is provided for the data tables.

ANSWERS TO CALCULATIONS These calculations are quite straightforward. The results should be very close to the accepted values given.

ANSWERS TO QUESTIONS IN THE MANUAL 1.

Define density, and give an example, using SI units. Ans: Density is defined as mass per unit volume. Example: The mean density of Earth is 5517 . g cm3 .

Calculate the volume of 1.00 g of pure, air-free water at atmospheric pressure and at the following temperatures. Make these calculations to four significant figures. (a) 2°C 22°C (b) Ans: Use Table 28.1 (a) By visual extrapolation the density of water at 2°C is above 0.9999 so the volume is 0.9999 ml (or cm3 ); (b) 0.9978 ml (or cm3 )

State the maximum density of pure water and give the temperature at which this maximum density occurs. . g cm3 , which occurs at 3.98°C. Ans: Maximum density = 10000

109

Determine the mass of a solid that has a volume of 6.39 cm3 and a density of 5.4 g/cm3. Be sure that your answer contains the correct number of significant digits. Ans: Mass = d × V = 6.39 cm3 × 5.4 g cm3 = 34 g . This answer is given to two significant digits.

Determine the volume of a liquid that has a mass of 24.6 g and a density of 0.82 g/cm3. Be sure the answer has the correct number of significant digits. Ans: Volume = m d = 24.6 g 0.82 g cm3 = 30 cm3 . Again, to two significant digits.

Explain how the volume of the human body could be determined experimentally. Ans: The volume of the human body can be determined by submerging the body in a liquid, collecting the liquid displaced, and then measuring the displaced volume.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Why does the density of water vary with temperature? Ans: A brief answer the student might give is that the density of water is a function of the mass of the water molecules and the spacing between them. The spacing is a function of temperature and becomes a minimum at 3.8°C.

How can density be helpful in identifying a pure metal from an alloy? Ans: The density of a pure metal is a definite physical constant. An alloy will have a slightly different density. A check on melting points will give further evidence.

Gold has a density of 19.32 g cm3 . A basketball has a volume of approximately 7 × 103 cm3 . Determine the mass of a solid gold basketball. Mass = d × V = 19.32 g cm3 × 7 × 103 cm3 = 135 . × 105 g = 135 . × 102 kg Ans: 2 135 . × 10 kg = 297 lb

110

Experiment 29

Oxygen INTRODUCTION If this is the first chemistry experiment, outline for students the special safety precautions for using and handling chemicals and equipment in the laboratory. You may want to review the safety rules at the beginning of the Laboratory Guide and present additional ones that pertain to your particular laboratory setup. In this experiment the instructor should dispense the manganese dioxide, because many students tend to use too much. In fact, some instructors keep a prepared mixture of KClO3 and MnO2 on a side laboratory table in test tubes ready for use. If this is the students’ first experience with the Bunsen burner, give instructions for lighting, adjusting, and using it. Mention where the flame has the highest temperature. Also emphasize the precautions necessary when heating the KClO3––MnO2 mixture. The answers to the procedures are the observations to the tests done with oxygen.

TESTS WITH OXYGEN 1.

The flame should rekindle and burn brightly.

Initial flame is blue and very low. In oxygen the flame turns yellow and increases in size.

Tip of wire will glow much more brightly.

The reaction in the oxygen produces a burst of glowing sparks like a sparkler on the 4th of July.

ANSWERS TO QUESTIONS IN THE MANUAL 1.

What do you think holds the water in the bottles that, after being filled, are inverted in the pneumatic trough? Ans: Atmospheric pressure holds the water in the collecting bottles.

111

From the results of Tests 1 and 2 in this experiment, what statement could be made about the effect of oxygen on the process of combustion? Ans: The pure oxygen supports and enhances combustion because normal air is only about 40% oxygen.

When sulfur is oxidized (combined with oxygen), of what chemical elements is the product composed? Ans: Sulfur and oxygen

In Test 4, the sulfur does not become part of the final product. Comparing Tests 3 and 4 of this experiment, what was the purpose of the sulfur in Test 4? Ans: The sulfur retains the heat and allows the iron to reach its kindling temperature so the iron actually burns.

In Test 4, what happened to the iron wire? Of what chemical elements was the product of the reaction composed? Ans: The iron was oxidized. Iron and oxygen

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Why is it possible to collect oxygen by the displacement of water? Ans: Oxygen does not react with water or dissolve in water.

How many grams of oxygen can be obtained from 100 g of KClO3? K

Ans:

39 + 35+ 48 = 122 100 x = 122 48 x = 39 g

Write the electronic configuration for oxygen. 1s 2 2 s 2 2 p 6 3s 2 3 p 4 Ans:

Give the Period and Group number for oxygen. Ans: Period 2, Group 6A

Oxygen is the most abundant element in Earth’s crust. Refer to Section 11.3 in the textbook and look up the percentage (by weight) of oxygen in the Earth’s crust. Ans: From the pie chart (Figure 11.9 on page 288) this percentage is 46.5 percent

112

Experiment 30

Percentage of Oxygen in Potassium Chlorate INTRODUCTION The procedure of heating a test sample to a constant weight is often used by scientists in determining the composition of compounds. In this experiment, when the weight difference between two successive weighings is 0.01 g or less, it may be assumed that all the oxygen has been driven from the potassium chlorate. Caution students to heat the test tube gently and evenly, being careful not to generate any hot spots on the test tube. To do this the Bunsen burner may have to be slowly moved back and forth along its length. The following data are typical for this experiment. DATA TABLE 30.1 Weight of test tube and MnO2

21.45 g

Weight of test tube, MnO2, and KClO3 before heating

20.24 g

Weight of test tube and residue after heating

20.99 g

COMPUTATIONS Weight of KClO3

0.46 g

Weight of oxygen driven off

38.0%

Percentage of oxygen in KClO3

21.45 g

From the formula KClO3 and the known atomic weights of its elements, calculate the theoretical percentage of oxygen in potassium chlorate. Compare this with the value obtained in your experiment by computing percent error; the percent error should be small. Theoretical percentage of oxygen in KClO3 Percent error in this data

39.2% 3.1%

113

ANSWERS TO QUESTIONS IN THE MANUAL 1.

What is the purpose of the manganese dioxide in this experiment? Ans: The manganese dioxide is a catalyst. It speeds up the reaction in the experiment but does not enter into the reaction products themselves.

What is the theoretical percentage of Cl in KClO3? Show your calculations. K

Ans:

39 35 48 122 + + = 122 122 122 122 35 × 100 = 29% for chlorine 122

What is the theoretical percentage of K in KClO3? Show your calculations. K

Ans:

39 35 48 122 + + = 122 122 122 122 39 × 100 = 32% for potassium 122

What do you think is the greatest source of error in this experiment? Why? Ans: Most students with poor results fail to remove all the oxygen from the KClO3. Some students are not careful in using and reading the balance.

Balance the chemical equation: ________ KClO → _____ KCl + ________ O2 Ans:

2 KClO 3 → 2 KCl + 3O 2

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What is the residue remaining in the test tube after this experiment is completed? Ans: Potassium chloride

Determine the percentage composition of KCl. K

Ans:

114

39 35 74 + = 74 74 74

b b

39 × 100 = 52.7% for potassium 74 35 × 100 = 47.3% for chlorine 74

Experiment 31

Percentage of Oxygen and Nitrogen in the Air INTRODUCTION This experiment will give the percentage of nitrogen and oxygen in the air within 5 percent of accepted values, provided there are no air leaks in the system. A good test for leaks in the system can be made as follows: With no heat applied to the combustion tube, pour 100 mL of water into bottle B1. This will force 100 mL of air through the system that will be collected in bottle B2, providing there are no leaks in the system. Carefully measure the volume of displaced gas in B2 to check this. Caution: Do not allow students to remove the copper turnings from the combustion tube. They tend to burn their fingers and consequently drop one end of the tube from the clamp. The tube then falls on the table and breaks. Let the tubes cool completely before the copper turnings residue is pushed carefully from the tubes with a wooden dowel rod. The following data are typical for this experiment. DATA TABLE 31.1 First Trial

Second Trial

Number of mL of nitrogen

240

238

Number of mL of air forced through

300

Percentage of nitrogen in air

80%

79.3%

Percentage of oxygen in air

20%

20.7%

COMPUTATIONS

ANSWERS TO QUESTIONS IN THE MANUAL 1.

Why must the air be forced slowly through the combustion tube? Ans: The air was forced slowly through the combustion tube to allow sufficient time for the oxygen in the air to react with the copper so that only nitrogen passes on into the collection bottle. 115

If part of the 300 mL of water were lost by an overflow of the thistle tube, how would the results of the experiment be affected? Ans: If part of the 300 mL of water input was lost, then 300 mL of air would not be forced through the combustion tube, and the experimental values for the percentage of nitrogen would be too low.

If an air leak developed at the right end of the combustion tube, how would the results of the experiment be affected? Ans: A leak at the right end of the combustion tube would result in a loss in the amount of nitrogen collected, which would give a lower percentage of nitrogen and a correspondingly greater amount in the percentage of oxygen.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Which element, nitrogen or oxygen, has the greatest weight in a sample of normal air? Ans: Nitrogen has the greatest weight.

Calculate the percentage composition of copper oxide (CuO). Cu O Ans: . . 635 16 79.5 635 + = × 100 = 79.9% copper 79.5 79.5 79.5 79.5 16 × 100 = 201% . oxygen 79.5

Why was bottle B2, containing the collected nitrogen, covered and then placed upright on the table? Ans: To prevent any further loss of water left in the bottle after the 300 mL or air has been processed.

116

Experiment 32

An Exothermic Chemical Reaction INTRODUCTION The amount of heat given off during a chemical reaction is directly proportional to the concentration of the reacting solutions. This experiment is written to illustrate this and to provide students with a simple hands-on method for understanding the exothermic concept. The calorimeter most widely used is the double-walled typed with a heavy, spun-aluminum inner cup. If this is not available, construct a coffee-cup calorimeter using two polystyrene cups. Place one cup inside the other, and place both in a beaker that will hold them firmly. If plastic cups are not available, use glass beakers. Place one small beaker inside a larger beaker insulated from one another by crumpled paper or some other insulating material. No specific data are provided for this experiment. The slope of the graph should be . °C ml . approximately 010

ANSWERS TO QUESTIONS IN THE MANUAL 1.

What does the graph indicate concerning the relationship between percent concentration of solution and the changes in temperature? Ans: The change in temperature is directly proportional to the percent concentration.

Explain the relationship between the heat released by the chemical reaction and the change in temperature. Ans: The heat released is directly proportional to the change in temperature.

Explain the slope of your graph. Ans: The slope of the graph is a straight line indicating that the change in temperature is directly proportional to the concentration.

How would the slope of your graph change if the volume of the solutions were doubled? Explain your answer. Ans: The slope will be approximately the same. The temperature change will be about the same. The amount of heat released will be twice as much because the volume was doubled.

117

Assume that the chemical reaction between the NaOH and the HC1 is complete. (a) With the 100% standard solution, how many sodium ions react with how many chlorine ions? (b) With the 25% solution, how many water molecules were formed? Ans: For the 2.0 M standard solutions used here: (a) Avogadro’s number × 2 that is 12.046 × 1023 sodium ions and the same number of chlorine ions. 12.046 × 1023 = 3.0115 × 1023 water molecules formed (b) 4

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Define an exothermic reaction. Give an example. Ans: An exothermic reaction is one that gives off heat. The combustion of methane (NH4) is an example.

Is the burning of a match an exothermic reaction? Ans: Yes.

Is an activation energy required to obtain the reaction in studied in this experiment? Ans: No, the reaction takes place spontaneously when the two solutions are mixed at room temperature.

118

Experiment 33

Avogadro’s Number INTRODUCTION If the classroom teacher has not lectured on Avogadro’s number, then briefly explain it at the beginning of the laboratory class. You might give a quick review of the introduction to the experiment and answer any students’ questions. Caution students to be careful in washing and handling the copper anode and cathode. The mass difference must be measured as accurately as possible. Also, the electric current must be kept constant. Keep the solution well mixed at all times, and adjust the rheostat for any voltage fluctuations to maintain a constant current reading on the milliameter. The following data are above average but can be obtained if the student is careful in determining the loss and gain in mass of the two copper electrodes. SAMPLE DATA DATA TABLE 33.1 1. Current

I = 1.92 A

2. Time

t = 1800 s

3. Loss of mass of anode (original mass – final mass)

mlost = 1.12 g

4. Gain of mass of cathode (final mass – original mass)

mgained = 1.14 g

5. Average mass transferred maverage =

mlost + mgained 2

maverage = 1.13 g

ANSWERS TO CALCULATIONS 1.

Determine the amount of flow of charge q. q = It q = _________ coulombs Ans:

q = I × t = 192 . C s × 1800 s = 3456 C = 3.46 × 103 C C Remember A = (units) s

119

Determine the number of electrons ne flowing through the solution, where q ne = −19 16 . × 10 coulomb ne = _________ electrons Ans:

−19

16 . × 10 coulomb 3 3.46 × 10 C ne = = 2.16 × 1022 electrons 1.60 × 10 −19 C

Determine the number of copper atoms that were transferred from the anode to the cathode. The charge carried by each copper atom is +2. n N c = no. of copper atoms transferred = e 2 N c = _________ electrons This is the number of copper atoms in the amount of mass transferred between the anode and cathode ( maverage ). Using this, we can set up a ratio of this mass to the mass of one gram mole of copper (63.54 g) and set this equal to the ratio of copper atoms transferred to Avogadro’s number: maverage N c = m1 mole No Ans:

ne =

ne 2 2.16 × 1022 nc = = 1.08 × 1022 copper atoms 2 nc =

Determine the number of copper atoms N o in 1 mole of copper. This will be equal to Avogadro’s number m N o = 1 mole N c maverage N o = __________

Ans:

There are 63.54 g of copper in 1 mole so the following ratio can be set up. 6354 . g: 114 . : : N o : 1.08 × 1022

g e

114 . N o = 6354 . × 108 . × 1023

N o = 6.02 × 1023 Note: this result is Avogadro’s number.

120

Determine your percent error using N0 = 6.022 × 1023 atoms/mole as the accepted value. __________ % Ans:

% Error =

6.02 × 1023 − 6.02 × 1023 6.02 × 1023

× 100 = 0%

For this data there is no error. Usual errors of around 6–8% are, however, common in this experiment.

ANSWERS TO QUESTIONS IN THE MANUAL 1.

Define one mole, and give an example. Ans: One mole is the amount of substance containing Avogadro’s number of any kind of chemical unit. For example, one gram molecular weight of water is 18 g.

State the numerical value of Avogadro’s number. 6.022 × 1023 Ans:

How many sodium ions are there in one mole of sodium chloride (NaCl)? How many chloride ions? Avogadro’s number Ans: (a) (b) Avogadro’s number

How many molecules are there in 47.8 grams of CuSO4? 47.8 × 6.02 × 1023 = 180 . × 1023 molecules Ans: 160

What do you think is the greatest source of error in this experiment? Why? Ans: The greatest source of error is probably in the determination of the loss and gain in mass of the two copper electrodes.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

State the periodic law of chemistry that is used to set up the periodic table. Ans: The properties of elements are periodic functions of the atomic number.

Define one gram atomic weight. Ans: The mass in grams of one mole of the naturally occurring atoms of an element.

Define electrolysis. Ans: The separation of a compound into simpler substances through the use of an electric current.

121

How many oxygen atoms are there in one mole of copper sulfate? Ans: There are 4 × 6.02 × 1023 atoms of oxygen in one mole of copper sulfate, because there are 4 oxygen atoms in each copper sulfate molecule.

How many hydrogen atoms are there in 18.0145 g water? Ans: There are 2 × 6.02 × 1023 atoms of hydrogen in one mole or 18.0145 g of H2O because there are 2 hydrogen atoms in each water molecule.

122

Experiment 34

Molecular Structure INTRODUCTION Although the introduction to this experiment discusses chemical bonding, it may be worthwhile to explain covalent and ionic bonding, again especially if students have not yet learned about chemical bonding in the lecture class. Also explain the concepts of isomers and how the models should be assembled and dismantled. A demonstration showing how to assemble and dismantle a model of CH4 is helpful. This experiment uses a molecular-models kit to construct models of molecules, so no specific data are provided.

ANSWERS TO QUESTIONS IN THE MANUAL 1.

What do the “dots” around and between the chemical symbols in the drawings in this experiment represent? Ans: The “dots” represent individual valence electrons.

What do the single, double, and triple “bars” between the chemical symbols in the drawing represent? Ans: The single “bar” represents the sharing or 1 pair, that is 2 electrons. The double “bar” represents the sharing or 2 pair, that is 4 electrons. The triple “bar” represents the sharing or 3 pair, that is 6 electrons.

How many valence electrons are associated with: a hydrogen atom a carbon atom a nitrogen atom a chlorine atom Ans: 1 valence electron 4 valence electrons 5 valence electrons 7 valence electrons

123

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Define covalent bond. Give an example. Ans: Covalent bonding is bonding in which atoms share electrons. H | O—H Example: H2O

What are isomers? Give an example. Ans: Isomers are molecules that have the same molecular formula but a different structure or arrangement of atoms, for example, n-butane and isobutane.

Distinguish between single, double, and triple bonds. Ans:

124

A single bond: Cl2

Cl — Cl

A double bond: CO2

O—C—O

A triple bond: N2

N— —N

Experiment 35

Solutions and Solubility INTRODUCTION This experiment is not very exciting for many students, so if the equipment is available, give a demonstration showing the Brownian movement or the Tyndall effect. If you demonstrate the Tyndall effect, you will have to explain a colloidal suspension, unless the lecture-class instructor has already introduced this concept. Careful measurement of temperature is very important in this experiment and care must also be taken not to spill solution when pouring it into the evaporating dish. Exact measurements of mass are also critical.

ANSWERS TO QUESTIONS IN THE MANUAL 1.

Define the term solution. Give an example. Ans: A solution is a homogeneous mixture of two or more substances—for example, sugar and water.

What is a saturated solution? Ans: A saturated solution is a solution in which the dissolved and undissolved solutes are in equilibrium.

Distinguish between solvent and solute. Ans: The solvent is the dissolving medium of a solution. The solute is the dissolved substance in a solution.

Define solubility. Give an example. Ans: The solubility of a given solute is the amount of solute that will dissolve in a specified volume of solvent (at a given temperature) to produce a saturated solution.

Is solubility a function of temperature? Explain. Ans: Yes, solubility is a function of temperature. Increasing the temperature usually increases the solubility of solids in liquids because of increased molecular activity.

125

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What effect would stirring the mixture have on the rate of dissolving the solid in a liquid? Ans: Stirring increases the rate of dissolving.

What effect would powdering the solid have on the rate of dissolving the solid in a liquid? Ans: Powdering the solid increases the rate of dissolving because it increases the surface area of the solute in contact with the solvent.

126

Experiment 36

Pressure-Volume Relationship of Gases INTRODUCTION Modern thermodynamics deals with the properties of systems for the description of which temperature is an essential quantity. The zeroth law of thermodynamics states the existence of a fundamental concept that we call temperature. The law defines temperature. The law states that if a system is in thermal equilibrium with two other systems, then these two systems are also in thermal equilibrium with each other. The word zeroth is used because the law was formulated after the first and second laws were stated but is considered even more fundamental. This experiment is concerned with a constant-temperature system. What is the relationship between the volume and the absolute pressure of a gas held at constant temperature (Boyle’s law)? The data in Data Table 36.1 were obtained using the apparatus shown in Figure 36.1in the Laboratory Guide.

127

SAMPLE DATA DATA TABLE 36.1 D = 2.40 cm

A= π r2 =

Mass M (kg)

Force F = mg (N)

Pmeas =

F A

(Pa)

Pabs = Pmeas + Patm (Pa)

Volume Vcc (cc)

Volume V (m3)

(Pa ⋅ m3 )

1.01

18.2

1.84

1.00

9.80

2.17

1.23

16.0

1.97

2.00

19.60

4.34

1.44

14.1

2.03

3.00

29.40

6.50

1.66

11.2

1.86

4.00

39.20

8.68

1.88

9.9

1.86

5.00

49.00

10.80

2.09

8.7

1.82

6.00

58.80

13.02

2.31

7.6

1.76

7.00

68.60

15.20

2.53

7.0

1.77

Patm = 1.013 × 105 Pa

128

4.52 cm2 = 4.52 × 10–4 m2 Pabs × V

(Pabs × V) av = 186 × 1010 Pa m3 % Difference in data = 11.3%

3.00 2.90

Pressure vs

1 for a gas at constant temperatur e volume

2.80 2.70 2.60 2.50 2.40 2.30 2.20 2.10 2.00 1.90 1.80 1.70 1.60 1.50 1.40 1.30 1.20 1.10 1.00

(

Volume m 3 × 10 6

)

129

ANSWERS TO QUESTIONS IN THE MANUAL 1.

Explain the shape of the curve you have plotted. Ans: The graph of P versus 1/V is a straight line because pressure is inversely proportional to the volume when the temperature is held constant. The reciprocal of the volume thus shows a direct relationship to the pressure.

State the relationship between the volume and pressure of a gas when the temperature is held constant (Boyle’s law). Ans: When the temperature is held constant, the pressure of a gas is inversely proportional to the volume.

Do the data taken in the experiment support the idea that the pressure times the volume for an ideal gas is constant when the temperature remains at the same temperature? Ans: Yes. See the last column in the data table.

The temperature of a gas is increased while the volume is held constant. Does the pressure decrease, increase, or remain the same? Explain. Ans: When the volume is held constant, the pressure of a gas is directly proportional to the absolute temperature. An increase in temperature increases the average kinetic energy of the molecules as the pressure is increased.

A volume of gas is decreased by half while the temperature remains constant. How much does the gas pressure change? Ans: The new pressure will be twice as great.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

When the temperature is constant, does the product of volume and pressure hold for any substance? Explain. Ans: No. The relationship holds only for ideal gases but it is a very good approximation for many real gases.

A sample of gas has a volume of 100 cm3 at a pressure of 80 cm of mercury and a temperature of 27°C. What would be the volume if the temperature remains the same and the pressure increases to 90 cm of mercury? 80 80 × 100 cm3 = 90 × V V= × 100 cm3 = 88.9 cm3 = 89 cm3 Ans: 90

Name the type of curve obtained in a graph of pressure versus volume. Ans: Hyperbola

Which temperature scale must be used in the ideal gas formula? Ans: If the pressure of a gas is held constant, the volume of the gas is directly proportional to its absolute temperature, so the Kelvin temperature scale must be used.

130

Experiment 37

Chemical Qualitative Analysis INTRODUCTION The objective of Experiments 37 and 38 is to provide students with sufficient experience to distinguish between qualitative and quantitative analyses and to see how each analysis is accomplished. It is best to give an unknown sample containing only one or two ions. Not obtaining a positive result makes students doubt whether they did the tests correctly. Having all three ions present in the unknown sample gives the same results as the trial run and appears to some students as a repeat performance of the analysis. This is probably the first experiment in which students will be handling chemicals that can harm their body and clothing. Inform them about the proper care and handling of reagents, showing them how to open, hold, and pour acids from reagent bottles. If not informed, many students lay the bottle stopper on their book, papers, or the lab table. The procedure and results for this experiment are shown in the flow diagram. No additional data is provided.

ANSWERS TO QUESTIONS IN THE MANUAL 1.

What is the purpose of qualitative analysis? Ans: Quantitative analysis is used to determine which chemical elements are present in a specific physical sample.

How is any lead chloride in the sample removed in step 2? Ans: Any lead chloride present is dissolved in the hot water that is poured through the filter paper in this step and is carried away as part of the filtrate.

131

What are the final indicators that show the presence of the following elements. Pb _______________________________________________________________________ Hg _______________________________________________________________________ Ag _______________________________________________________________________ Ans:

Pb—A yellow cloudy solution after step 3 shows the presence of lead. Hg—A black precipitate in the filter paper after step 4 shows the presence of mercury. Ag—A change from clear to cloudy (white) of the filtrate after step 5 indicates the pressure of silver.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Define qualitative analysis. Ans: Qualitative analysis is the systematic effort to determine what elements are present in a sample of unknown constituency.

Distinguish between a precipitate and a filtrate. Ans: A precipitate is usually a solid that separates from a solution as a result of some physical or chemical change and can be removed by filtration. A filtrate is the liquid that has passed through a filter.

How was the lead chloride separated from chlorides of lead and mercury? Ans: By adding hot water to the chlorides which dissolves only the lead chloride, if it is present.

How were the silver ions separated from the lead ions? Ans: The silver ions were separated from the lead ions by adding ammonium hydroxide to the ions.

Describe how a 1.0 M solution of hydrochloric acid is prepared. Ans: One mole of hydrochloric acid, HCl, has a mass of 36 g. Adding this amount of HCl to enough water to make exactly 1 L of solution produces a 1.0 M solution.

132

Experiment 38

Chemical Quantitative Analysis (Volumetric) INTRODUCTION The introduction for this experiment is given in Experiment 37. In Procedure 2, 1.0 M HCl and 0.5 M NaOH were used for the data given in Table 38.2. SAMPLE DATA DATA TABLE 38.1 Volume of vinegar Volume of sodium hydroxide

15 mL 10

COMPUTATIONS Weight of volume of vinegar

15 g

Weight of acetic acid present

0.6

Percentage of acetic acid

133

SAMPLE DATA DATA TABLE 38.2 Volume of hydrochloric acid

15 mL

Volume of sodium hydroxide solution

COMPUTATIONS Weight of volume of NaOH solution

15 g

Weight of NaOH present

1.2

Percentage of NaOH

ANSWERS TO QUESTIONS IN THE MANUAL 1.

Find the number of grams of KCl in 500 mL of a 3 molar solution. K

Ans:

39 + 35 = 74 g A 1.0 M solution will contain 74 g/L solution. A 3.0 M solution will contain 3 × 74 or 222 g/L solution. So 500 mL or 1/2 L will contain 222/2 = 111 g of KCl per 500 mL of solution.

What would be the molarity of an NaCl solution if 250 g of salt is used to make 5 L of solution? Na

Ans:

23 + 35 = 58 g (1 gram-molecular weight) 250 g of solute (NaCl)/5 L solution is the same as 50 g of solute (NaCl) in a single liter of solution. The molarity of this sample would be: 50/58, equal to 0.862 M.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Define one gram molecular weight. Give an example. Ans: The mass of a molecular substance in grams equal to the sum of all the molecular weights of the elements in the substance expressed in grams. Example: For water (H2O) 2 1 + 1 16 = 18 g

bg b g

Define molarity. Give an example. Ans: The concentration of a solution expressed in moles of solute per liter of solution. One . = 58.5 has a mass of 58.5 g. This amount of NaCl mole of NaCl 1 23 + 1 355 dissolved in enough water to make exactly 1 L of solution produces a 1.0 M solution.

b g b g

134

Explain how to make a 1.0 M solution of H2SO4. Ans: One mole of H2SO4 2 1 + 1 32 + 4 16 = 98 has a mass of 98 g. Dissolving this amount of H2SO4 in enough water to make exactly 1 L of solution gives a 1.0 M solution.

bg b g b g

Experiment 39

Kepler’s Law INTRODUCTION Although this experiment pertains to Kepler’s law of equal areas, the instructor should review Kepler’s other two laws of planetary motion. Also explain how the drawing at the end of the experiment was obtained, mentioning that it is an exaggerated ellipse representing a planetary orbit with the Sun at one focus. The actual path of a planet is more nearly a circle. The five triangles for Figure 39.2 in the Laboratory Guide were drawn and data for the experiment are given below. Suggest to your students that they begin with triangle number 1 as shown in the sample completed drawing of Figure 39.2/A, and number the other triangles as illustrated. This will make it easier for the grader to check the student’s experiment. The sample data provided show the accuracy that can be obtained in this experiment if is done carefully. It is very important to remind students to construct the perpendicular exactly 90° to the longest line (emphasize longest line), not at some angle other than 90°. Many students ignore this instruction and obtain very poor results. Triangle 6 Triangle 1

Triangle 5

Triangle 2 Triangle 3

Triangle 4

A Drawing showing the elliptical path of a revolving body with one possible set of triangles drawn in.

135

SAMPLE DATA DATA TABLE 39.1 Area of Triangle

W ×h cm2 2

Triangle

Width W (cm)

Height h (cm)

09.40

3.20

15.0

15.25

1.90

14.5

16.60

1.75

14.5

16.20

1.85

15.0

12.10

2.45

14.8

07.90

3.75

14.8

Average area =

14.8

% difference =

3.0%

e j

ANSWERS TO QUESTIONS 1.

What would happen if both pins in Procedure 1 were located at the same point, that is, right next to each other actually touching? (Try it if you need to, so that you can see what the result would be.) Ans: The line traced out will be a circle.

In Figure 39.2 the time interval between the dots is 0.10 seconds. What is the overall period for one swing of the pendulum bob that was photographed to make this figure? . = 185 . . Ans: There are 18 spaces between the dots so: 18 × 0105

What is the period (time for one complete revolution) for the planet Earth to travel around the Sun on its elliptical path? Ans: 365 days

If Fig. 39.2 represents a planet with the same period as Earth, how long does it take for the planet to move from one dot to the next? Ans: Time = 365 days/18 dots = 20.3 days.

Remembering that each set of dots is separated by the same time interval, where on its elliptical path will the planet be traveling the fastest? Ans: A planet will travel at its greatest speed when it is closest to the Sun where the dots are farthest apart.

136

If Earth is closest to the Sun in January of each year, what can we say about the speed of Earth in its orbit around the Sun at this time of year? Ans: In January the speed of Earth in its orbit around the Sun will be the greatest that it is all year.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Earth is closest to the Sun in January and farthest from the Sun in July. Compare the number of seconds in an Earth day for these two months. Ans: The number of seconds in an Earth day is greater in January, because Earth is closer to the Sun in January and traveling at a greater orbital speed. Greater orbital speed causes Earth to travel a greater orbital distance. Thus Earth must rotate through a greater angle for one complete day to occur, that is, from 12 noon to 12 noon.

A pencil, two thumbtacks, and a loop of string with a circumference of 29 cm are used to draw an ellipse. How does the shape of the ellipse change as the distance between the two thumbtacks is increased? Ans: The ellipse becomes more egg-shaped as the distance between the two tacks is increased, that is, the eccentricity increases as the distance between the two tacks increases.

The average distance between Earth and the Sun is 9.3 × 107 mi. Assume a circular orbit for Earth and calculate the distance Earth travels in 1 year. Circumference = 2π r = 2π × 9.3 × 107 mi Ans: d = 58 . × 108 mi

Use the distance calculated in Question 3 and determine Earth’s speed in mi/h. d 5.8 × 108 mi Ans: v= = = 6.7 × 104 mi h 24 hours t 365 days × 1 day

137

Experiment 40

Stars and Their Apparent Motions INTRODUCTION This is a long experiment, and it is difficult for some students. The instructor should go through the discussion part of the experiment step by step with the entire class, especially the setting of the celestial sphere. Explain how this sphere is set for a latitude, a day, and a time of day. Have the students set the sphere as you give the directions, answering questions as you proceed. Circulate in the laboratory, checking to see if students have the sphere set correctly. Pay particular attention to the angle for the correct altitude of the pole star. Always have students count the number of degrees to make sure the sphere is set correctly. Many students make an error in setting the celestial sphere for this angle and get incorrect answers when using it to solve problems. The following data are for October 26. If you want answers for the experiment to match the dots given here, tell your students to use this same date. Many students will have trouble with Questions 4, 5, and 6 in Procedure 2. For this date (October 26) the answers will be easier to obtain if the star Vega is used instead of Sirius.

PROCEDURE 1 1.

Declination of the Sun Right ascension of the Sun Altitude of the Sun Altitude of Polaris

7° S 13 hours 83° 0°

Altitude of the Sun Altitude of Polaris at this time

22° above the western horizon 0°

Altitude of the Sun Altitude of Polaris at this time

12° below the horizon 90°

Altitude of the Sun Altitude of Polaris at this time

12° below the horizon 90°

Altitude of the Sun Altitude of Polaris at this time Time of sunrise Time of sunset Hours of daylight Hours of darkness

39° 39° 7:30 A.M. 5:30 P.M. 11 13 139

Daylight hours at equator Daylight hours at north pole Daylight hours at south pole Apparition Occultation 6.

12 0 24 51°N to 90°N 51°S to 90°S

Declination of the Sun (June 21) Declination of the Sun (December 21) Altitude of the Sun on June 21 Degrees east of due south for sunrise on June 21 Degrees west of due south for sunset on June 21 Altitude of the Sun on December 21 Degrees east of due south for sunrise on December 21 Degrees west of due south for sunset on December 21

23.5°N 23.5°S 74.5° 108° 108° 28.5° 56° 56°

PROCEDURE 2 1.

See the drawing on the following page for this answer.

Great Nebula in Andromeda on the overhead meridian

Declination Right ascension

Time Sirius will rise

11:20 P.M.

Sirius on the overhead meridian at

4:45 A.M.

Altitude of Sirius

10:42 P.M. +41° Zero hours 42 min

35°

Cassiopeia Polaris

Little Dipper 39° Big Dipper

Northern horizon

Diagram for stars as seen on October 10 at 10:00 P.M. from 40°N latitude 140

ANSWERS TO QUESTIONS IN THE MANUAL 1.

How does the declination of the Sun vary over 1 year? Ans: The declination of the Sun varies from 23.5°N to 23.5°S back to 23.5°N during 1 year. The starting point for the answer is optional, but you must start at 23.5° or less.

Where on the ecliptic is the vernal equinox located? Ans: The vernal equinox is located on the celestial sphere where the Sun crosses the celestial equator going from south to north.

How does the altitude of the North Star vary for an observer located at the equator? At 39°N? Ans: For an observer at the equator, the altitude of Polaris is 0° at all times. For an observer at 39°N, the altitude of Polaris is 39° at all times.

Why was the globe rotated westward when positioning the celestial sphere to portray a P.M. (afternoon) local time? Ans: The globe was rotated westward to portray a P.M. local time because the Sun appears to travel westward across the sky as the day progresses.

Does the Sun ever set north of due west for an observer located at Washington, D.C. (39°N)? What time of year would this occur? Ans: Yes, during the winter months of each year.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

From your local latitude, what is the declination of a star that appears on the southern horizon (0° altitude)? Ans: The answer depends on the latitude as perceived by the observer. The star will be 90° south of the observer’s position. Example: If the observer is at 40°N, the declination of the star will be 90° south of 40°N. The answer is 50°S.

What is the maximum altitude of the Sun for an observer located at the north pole (90°N)? On what date will the Sun be observed at the maximum altitude? Ans: Maximum altitude is 23.5°, which will be observed at the time of summer solstice (June 21 ± 2 days).

It is July and you are in your hometown. As you travel northward toward the north pole, how does the number of daylight hours vary? How does the number of daylight hours vary as you travel southward toward the south pole? Ans: The number of daylight hours increases as you travel northward. As you travel southward the number of daylight hours decreases.

Answer Question 3 using January as the date. Ans: The number of daylight hours decreases as you travel northward. The daylight hours increase as you travel southward. 141

Sirius, the brightest star in the sky, is 8.7 ly from the Sun. Determine Sirius’ declination and right ascension. Ans: Declination: 16° 39’; right ascension: 6 h 43 min as found on the celestial sphere. The distance that the star is from the Sun does not affect this answer.

Sirius, the brightest star in the sky, is part of what constellation? Ans: Canis major as found on the celestial sphere or in stellar tables.

142

Experiment 41

Locating Stars in the Night Sky INTRODUCTION Looking at the night sky, even without a telescope, can be interesting and fun. Some students really enjoy pointing out the brightest stars and the shapes of constellations to their families and friends. This experiment uses star charts, a celestial sphere, and actual observation of the stars to teach students how to find their way around the night sky during one of the four seasons of the year.

EQUIPMENT AND PROCEDURE TIPS Procedures 1 and 2: The first element of this study involves the use of two star charts that show the position of the brightest stars in the night sky for observers at 40°N latitude. There are actually four sets of these paired star charts, one set for each season of the year. Choose the pair of star charts that best represents the sky for the current season of the year. The first star chart in each pair shows not only the brightest stars but also the outline of the constellations in which these stars are found. The second star chart in each pair shows only the stars. After studying the outline of the constellations and the names of the brightest stars on the first star chart, the student is to fill in from memory as much data as possible on the second chart, which shows only the stars. All four sets of seasonal star charts are provided in the Laboratory Guide, so the only additional equipment needed for this part of the experiment is a pencil. Procedure 3: After about 10 minutes of work from memory, the students are allowed to compare their work on the second chart with the first one showing the constellations and star names. This is a very important part of the experiment. It allows the students to immediately check on the accuracy and completeness of their knowledge of the star charts for the current season of the year and also reinforces that knowledge so that they will be more prepared to find stars in the night sky. Notice that the star charts can be used for any month by adjusting the observer’s viewing time as indicated at the bottom of each chart. This allows one chart to serve each season of the year. Procedure 4: A table of stars in the Laboratory Guide provides the names of eight of the brightest stars that are found in the night sky during each season of the year. After the names of the stars being studied during the chosen season of the year have been transferred to Data Table 41.1 in the Laboratory Guide, students should fill in the names of the appropriate constellations; then they should use a celestial sphere to find each star and determine its right ascension, declination, altitude, 143

zenith angle, and the approximate compass direction for the current date at 9:00 P.M. local solar time. The procedures for using a celestial sphere are explained in Experiment 40 and will not be repeated here. If a celestial sphere is not available, the right ascension and declination can be found from a star atlas or from the master list of stars provided in this Instructor’s Guide (Table B). The compass direction can be determined from the star charts by drawing a North––South, East––West coordinate system through the observer’s zenith point (the center of the circle on the star chart). A line can then be drawn from the center of the chart through the star and its compass direction determined by using a protractor. The altitude and zenith angle are not as easy to find without a celestial sphere because any polar projection map becomes greatly distorted toward the edges. The best you will be able to do is a rough estimate of these values. Remember that Polaris has an altitude of 40º if you are located at 40ºN latitude. This can give you some guidance as to the general altitude of other stars on the star charts. The student is asked to find information for the current date and for the same hour (9:00 P.M. local solar time) as the star charts indicate. This time may be adjusted to Standard Time if you wish, but the location of the stars will be within 0.5 hours (about 7º) of their correct locations no matter where you are located with respect to the nearest standard meridian. Therefore, the sky will appear much the same whether or not this correction is made. Daylight Savings Time, however, should be taken into account because this can cause about 15º of shift in the position of the stars. If you are on Daylight Savings Time, subtract 1 hour from the times indicated on the star charts. Procedure 5: Again, the master list of stars provided in this Instructor’s Guide (Table B) can be used to fill in Data Table 41.3 in the Laboratory Guide if a good star atlas is not available. You may wish to make several photocopies of Table B for in-class use. The right ascension and declination that are filled in during this step should be checked to see that the data in Data Table 41.1 in the Laboratory Guide was found correctly using the celestial sphere. If the values are not close, the student should recheck all the information for that star again. The data in both Data Tables 41.1 and 41.2 in the Laboratory Guide will help the students find the stars in the actual night sky in the next procedure. Note that all the stars used are second magnitude or brighter, so they will be the key stars observed when the students look up into a clear sky at night. The spectral class and the color of a star as seen in the night sky are correlated in Table A. You also may wish to provide a copy of this table for student use when they compare their observations in Data Table 41.3 with the data taken in Data Tables 41.1 and 41.2 in the Laboratory Guide.

Color Reddish Yellow White Blue-white

Table A Spectral Class K and M class stars G class stars A and F class stars O and B class stars

Procedure 6: The real fun of this experiment comes in applying all the information learned in Procedures 1–5 to understanding and observing the night sky. Data Table 41.3 in the Laboratory Guide provides space for recording each student’s actual observations. Data Table 41.2 in the Laboratory Guide should be used as a reference when the student goes outside to observe the night sky. This table, along with the star charts, gives the altitude and compass direction for each of the brightest stars. This information should be very helpful when actually locating these stars in the sky. After the stars have been located, their observed zenith angle and compass direction are recorded along with the apparent magnitude (brightness) and the observed color of each star. These colors are sometimes quite subtle, especially if there is haze or dust in the air. The best way to note the difference in color is to compare one star to another while observing them in the sky. Table A can be used to compare the observed color of each star to the color associated with its spectral class. 144

Procedure 7: A good source of data on planets currently observable in the night sky may be found in either Sky and Telescope or Astronomy. These magazines can be purchased at most newsstands or supermarkets. Since these magazines show data for one month in advance of their release dates, you should purchase a copy about 30 days before you plan to give out information to your students. Your local library or school library also may have copies of these magazines. Up-dodate star charts and tables show the position of the planets. The information is given near the center of both of these publications. The answers to Questions 1 and 2 at the end of Procedure 7 should be given to the students before they go outside to observe the night sky. They should provide the answers to Questions 3 and 4 from their actual observations. Table B The Twenty-Five Brightest Stars (as Seen from 40ºN Latitude) Apparent Spectral Right Ascension Declination Magnitude Luminosity* Class Star h min degrees Adhara 6 57 –28.9 1.6 9000 B Aldebaran 4 34 +16.5 0.9 150 K Algol 3 06 +40.8 2.1 V** 130 B Alnitak 5 39 –2.0 1.8 35000 O Altair 19 49 +8.8 0.8 10 A Antares Arcturus Bellatrix Betelgeuse Capella

16 14 5 5 5

28 14 24 54 14

–26.3 +19.4 +6.4 +7.4 +46.0

1.0 –0.1 1.6 0.4 V** 0.1

10000 110 3800 15000 140

M K B M G

Castor Deneb Dubhe Elnath Fomalhaut

7 20 11 5 22

33 40 02 24 56

+32.0 +45.2 +61.9 +28.6 –29.7

1.6 1.3 1.8 1.7 1.2

35 55000 150 1500 13

A A K B A

Mirfak Mirzam Polaris Pollux Procyon

3 6 2 7 7

22 21 03 43 37

+49.8 –17.9 +89.1 +28.1 +5.4

1.8 2.0 2.5 1.2 0.4

4700 7000 5500 33 8

F B F K F

Regulus Rigel Sirius Spica Vega

10 5 6 13 18

07 13 43 24 36

+12.1 –8.3 –16.7 –11.0 +38.8

1.4 0.2 –1.6 1.0 0.0

155 55000 23 1700 52

B B A B A

*Luminosity is expressed as a multiple of the absolute brightness of the Sun. **V indicates a variable magnitude star. Sample Data: Data taken for the date April 21 at 9:00 P.M.

145

SAMPLE DATA FOR SPRING Name of Star

DATA TABLE 41.1 Right Constellation Ascension Declination Altitude h

min

Zenith Angle

Compass Direction

Aldeberan

Bootes

+20º.0

45º

8ºS of E

Algol

Areiga

45º.0

32º

58º

38ºN of W

Altair

Gemini

+31º.0

55º

35º

6ºN of W

Capella

Ursa Minor

+89º.0

40º

50º

North

Deneb

Gemini

+28º.0

55º

35º

5ºS of W

Formaihaut Canis Minor

+5≤°.0

32º

58º

20ºS of W

Polaris

Leo

+13º.0

62º

28º

67ºS of W

Vega

Virgo

–10.5º

30º

60º

48ºS of E

Note: Altitude + zenith angle = 90° DATA TABLE 41.2 Name of Star

Right Ascension

Declination

Apparent Magnitude

Luminosity Compared to Sun

Spectral Class

min

1 Arcturus

+19.4º

–0.1

110

2 Capella

+46.0º

0.1

140

3 Castor

+32.0º

1.6

4 Polaris

+89.1º

2.5

5500

5 Pollux

+28.1º

1.2

6 Procyon

+5.4º

0.4

7 Regulus

+12.1º

1.4

155

8 Spica

–11.0º

1.0

1700

146

Name of Star

Constellation

DATA TABLE 41.3 Approximate Approximate Zenith Compass Angle Direction

Apparent Color of Star

Apparent Magnitude of Star

1 Arcturus Boötes

45º

10ºS of E

Reddish

1st

2 Capella

Auriga

60º

40ºN of W

Yellow

1st

3 Castor

Gemini

35º

10ºN of W

Blue-white

2nd

4 Polaris

Ursa Minor

50º

North

White

2nd

5 Pollux

Gemini

35º

5ºS of W

Reddish

1st

6 Procyon

Canis Minor

60º

25ºS of W

White

1st

7 Regulus

Leo

30º

70ºS of W

Blue-white

1st

8 Spica

Virgo

60º

50ºS of E

Blue-white

1st

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What is the brightest star in the sky other than the Sun? Ans: Sirius, which has an apparent magnitude of –1.6.

How do we specify the location of a star on the celestial sphere? Ans: We use two coordinates called right ascension and declination to specify a star’s position.

How do we specify the location of a star in the night sky? Ans: We use two coordinates, altitude or zenith angle, and compass direction.

Define altitude. Ans: The altitude of a star or other object in the sky is the angular distance above the horizon at which that star or object can be found at that date and time.

147

Experiment 42

Motions and Phases of the Moon INTRODUCTION This experiment will be easier for students to complete because they have now had experience with the celestial sphere. It is worthwhile to demonstrate the motion and phases of the moon using a light bulb for the Sun, an Earth globe, and a ball about one-fourth the diameter of the Earth globe for the Moon. The demonstration is more effective when performed in a darkened laboratory. This attracts the students’ attention, and they are unable to read or work on the experiment in the dark. The use of masking tape to represent the Sun and the Moon on the celestial sphere is a must. The students have a difficult time using only figures to keep track of the positions for the Sun and the Moon on the celestial sphere. The celestial globe does not fit perfectly in the cradle, so the answers obtained may not be all that precise. When grading, allow ±3° in angle measurements and ±12 min in time measurements.

PROCEDURE 1 DATA (a)

Altitude (Moon on overhead meridian)

27.5°

(b)

Altitude (Moon on overhead meridian) Local solar time the Moon rises Local solar time the Moon sets

51° 7 A.M. 5 P.M.

(c)

Altitude (Moon on overhead meridian) Local solar time the Moon rises Local solar time the Moon sets

74.5° 4:45 A.M. 7:15 P.M.

(d)

Altitude (Moon on overhead meridian) Local solar time the Moon rises Local solar time the Moon sets

51° 5:45 A.M. 6:15 P.M.

149

PROCEDURE 2 Washington, D.C. (39°N, 77°W) is the observer’s position. Assume that Washington is on Eastern Standard Time, rather than Daylight Saving Time. (a)

Date of new moon Altitude (Moon on overhead meridian)

September 23 51°

(b)

Date of first-quarter moon Local solar time first-quarter moon is on overhead meridian Standard time first-quarter moon is on overhead meridian Altitude of first-quarter moon when on overhead meridian Rising time (local solar) of first-quarter moon Rising time (standard) of first-quarter moon Setting time (local solar) of first-quarter moon Setting time (standard) of first-quarter moon

October 1 6 P.M. 6:08 P.M. 28° 1 P.M. 1:08 P.M. 10:30 P.M. 10:38 P.M.

(c)

With the passing of another 7 3/8 days, the Moon will be in full phase. Solve for the following information using previously learned facts and the celestial sphere. Date of full moon October 8 Local solar time full moon is on the overhead meridian 12 midnight Standard time full moon is on the overhead meridian 12:08 A.M. Altitude of full moon when on overhead meridian 58° Rising time (local solar) of full moon 5:20 P.M. Rising time (standard) of full moon 5:28 P.M. Setting time (local solar) of full moon 6:20 A.M. Setting time (standard) of full moon 6:28 A.M. Degrees east of due south at which full moon rises 87° Degrees west of due south at which full moon sets 87°

(d)

When the Moon has revolved to 270° east of the Sun, it will be in the last-quarter phase. Solve for the following information using previously learned facts and the celestial sphere. Date of last-quarter moon October 15 Local solar time last-quarter moon is on the meridian 6 A.M. Standard time last-quarter moon is on the meridian 6:08 A.M. Altitude of last-quarter moon when on overhead meridian 73° Rising time (local solar) of last-quarter moon 10:40 P.M. Oct. 14 Rising time (standard) of last-quarter moon 10:48 P.M. Oct. 14 Setting time (local solar) of last-quarter moon 1:20 P.M. Setting time (standard) of last-quarter moon 1:20 P.M.

ANSWERS TO QUESTIONS IN THE MANUAL 1.

150

During what month will the full moon achieve its maximum altitude, as observed from Washington, D.C. (39°N)? When will it have its minimum altitude? Ans: The full moon will be observed at maximum altitude during the month of December from latitude 39°N. Minimum altitude will be observed during the month of June.

Is it possible to see a waxing crescent moon on the overhead meridian at 10 A.M. local time? How about at 4 P.M. local time? Explain your answers. Ans: A waxing crescent moon cannot be observed on the overhead meridian at 10 A.M. local time but can be seen on the overhead meridian at 4 P.M. The waxing crescent moon begins as the new moon ends (12 noon local solar time), continues for 7 3/8 days, and ends at the time of the first-quarter phase of the moon. Because the new moon occurs at 12 noon local solar time, and the first quarter at 6 P.M., the waxing crescent can be seen on the overhead meridian between these two times.

In this experiment, what is the maximum possible error in degrees that could be made in determining the Moon’s altitude by assuming the Moon traveled in the same plane as the ecliptic? Ans: The maximum possible error would be plus or minus 5°.

Assuming the Moon actually travels in the same plane as the ecliptic, how many total lunar eclipses would occur each year? How many total solar eclipses? Ans: There would be a total of 12 lunar and 12 solar eclipses each year if the two planes were the same.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

For an observer at the equator, what will be the time the full moon rises on February 1? Ans: The full moon rises about 6 P.M. local solar time.

Will the full moon’s rising time for an observer at 40°N on February 1 be the same, earlier, or later than the rising time for an observer at the equator? Ans: The full moon will rise earlier for the observer at 40°N.

Is it possible for an observer at 28°N to see the full moon on the zenith? Why or why not? Ans: It is possible for an observer at 28°N to see the full moon on his or her zenith. The maximum northern latitude of the Moon is 23.5° + 5° or 28.5°.

What is the maximum altitude of the full moon as observed from Washington, D.C. (39°N)? How often will this maximum altitude occur? Ans: The maximum altitude will be 90° – [39° – (23.5° + 5°)] = 79.5° This will occur once every 18.6 years. The orbital plane of the Moon precesses and makes one complete cycle in 18.6 years. The 5° angle is added to 23.5° once every 18.6 years.

If the Moon revolved in the equatorial plane, what would be the maximum altitude of full moon for an observer at Washington, D.C.? How often would the maximum altitude occur? Ans: The maximum altitude would be 90° – 39° = 51°. Maximum altitude of the full moon would occur once every 29.5 days.

151

The average distance from Earth to the moon is approximately 238,000 mi. (a) Calculate the average orbital speed of the Moon in mi/h. Compare this with Earth’s orbital speed around the Sun. (b) Ans:

(a)

C = 2π r = 2π × 238,000 mi = 15 . × 106 mi d 1.5 × 106 mi v= = = 2.1 × 103 mi h 24 h t 29.5 days × 1 day

(b)

Refer to Additional Question 4 in Experiment 39 for the orbital speed of Earth. 6.7 × 104 mi h 2.1 × 103 mi h

= 32

Earth’s orbital speed is about 32 times greater than the Moon’s orbital speed.

152

Experiment 43

Observing the Phases of the Moon INTRODUCTION This experiment is done outside of class by observing the Moon over a period of several weeks to learn its phases and when it will be visible in these various phases. All observations will be made with the unaided eye, so telescopes and binoculars are not required. The observing sequence can begin at any time during the month, but the students must be told the dates of the next full- or new-moon phases so that they can plan their observations accordingly. Data Table 43.2 can be filled in at any time. It might be a good idea to have the students do this on the day that this assignment is first given so that they will have a reference that indicates the time when each phase of the Moon will be above the horizon. In this table the Moon is assumed to be above the horizon for 12 hours each day. In reality, the duration of time the Moon is up can vary by one or more hours depending on the season of the year. If you are observing from the northern hemisphere in the summertime, the new moon will be above the horizon longer than our assumed 12 hours because it is high in the sky with the Sun. This corresponds to the longer hours of daylight observed during the summer for residents of the northern hemisphere. In contrast, during the summer, the full moon will be low on the horizon and above the horizon less than 12 hours each day because its declination is 23.5°S during this season of the year. In the winter, the reverse is true. The duration of time that the Moon is actually above the horizon also varies with latitude. The rising and setting times in Data Table 43.1 are only a rough guide as to when the Moon actually will be on the observer’s horizon. Data Table 43.2 is to be completed outside of class and can only be filled in on days when the sky is clear enough for the Moon to be seen. You may want the students to record each date for about two-thirds of a month, indicating observational data as well as those dates on which no observation could be made because of the weather, or you can have students skip the dates when no observation was made, leaving room in the table for a full month of observations. The observing location, phase of the Moon, date and time of observation (using standard time as read from your everyday clock) must be recorded. The altitude may be difficult to estimate unless you have your students build a simple astrolabe from a piece of cardboard, a straw, a piece of string, and a weight such as a small nut or washer. (See Science Resources for Schools,* Vol. 3, No. 1.) Even if only a rough estimate is made, the student should be able to get to within about 10° of the correct altitude by using a simple hand-span calculation. (When your arm is held straight out, one hand-span covers about 15° of sky.) The student also must know the direction of due south in order to estimate the compass direction of the Moon at the time of observation. Here the degree markings on a homemade astrolabe may be helpful. An estimate to within 5 to 10 degrees is all that is normally expected for this reading. *Sponsored by the American Association for the Advancement of Science, 1333 H St., N.W., Washington, D.C. 20005.

153

The last column in Data Table 43.2 requires that the local solar time be calculated when the Moon will be (or was) on the observer’s overhead meridian. Using the fact that one hour passes for each 15° that the Moon is east or west of your overhead meridian, you must estimate the time the Moon will be on your overhead meridian. This correction is then added to or subtracted from the time of observation to give the standard time at which the Moon was or will be on your overhead meridian that day. This time cannot be compared directly with the information in Data Table 43.1 until it has been changed to local solar time. The difference between standard time and local solar time depends on where you are with respect to your nearest standard meridian. The standard time zones in the continental United States and the meridians that keep the mean times are Eastern Standard Time (EST) 75°W, Central Standard Time (CST) 90°W, Mountain Standard Time (MST) 105°W, and Pacific Standard Time (PST) 120°W. After the local longitude is known, we can then determine how many degrees east or west of the nearest standard meridian your observation point will be. The time correction is 4 minutes for each degree. The local solar time will be later if you are west of the nearest standard meridian and earlier if you are east of the nearest standard meridian. The sample data given below are for observations made from Zanesville, Ohio (40°N, 82°W). The time correction will be for the Eastern Standard Time zone. 82°W – 75°W = 7° 7° × 4 minutes = 28 minutes, or about 0.5 hours Since this observation site is west of the 75°W standard meridian, the local solar time (LST) will be later than the clock time (EST), and the estimated time for the Moon to be overhead will be 0.5 hours later than the EST (clock time) of the observation.

Phase

DATA TABLE 43.1 Moon on Overhead Meridian Moon Rises

Moon Sets

New moon

6:00 A.M.

12:00 noon

6:00 P.M.

Waxing crescent

9:00 A.M..*

3:00 P.M.*

9:00 P.M.*

First quarter

12:00 noon

6:00 P.M.

12:00 midnight

Waxing gibbous

3:00 P.M.*

9:00 P.M.*

3:00 A.M.*

Full moon

6:00 P.M.

12:00 midnight

6:00 A.M.

Waning gibbous

9:00 P.M.*

3:00 A.M.*

9:00 A.M.*

Third quarter

12:00 midnight

6:00 A.M.

12:00 noon

Waning crescent

3:00 A.M.*

9:00 A.M.*

3:00 P.M.*

*The times marked with an asterisk must be considered the center point of a ± 3 hour range because the crescent and gibbous phases of the Moon last for nearly a week, so the times for rising, being overhead, and setting can be within a six hour range.

154

Observing Location for this Record

DATA TABLE 43.2 Personal Observation Log (Sample data) Date/Time of Phase of Observation Altitude/ (Standard Time) the Moon Direction DST Observed of the Moon

Estimated LST Time When the Moon Was or Will Be Overhead*

Home

Full

6/14 / 10:30 P.M.

15° / 40°E of S

12:00 midnight

Home

Waning gibbous

6/15 / 10:30 P.M.

05° / 50°E of S

01:00 A.M.

—

Waning gibbous

6/16 / Cloudy

Work

Waning gibbous

6/17 / 7:30 A.M.

05° / 45°W of S

02:00 A.M.

—

Waning gibbous

6/18 / Cloudy

—

Waning gibbous

6/19 / Cloudy

Home

Waning gibbous

6/20 / 7:30 A.M.

35° / 35°W of S

04:30 A.M.

Home

Waning gibbous

6/21 / 7:30 A.M.

40° / 20°W of S

05:00 A.M.

Work

3rd quarter

6/22 / 7:30 A.M.

50° / 15°W of S

06:00 A.M.

Work

Waning crescent

6/23 / 7:30 A.M.

55° / South

07:00 A.M.

—

Waning crescent

6/24 / Cloudy

—

Waning crescent

6/25 / Cloudy

Work

Waning crescent

6/26 / 7:30 A.M.

55° / 65°E of S

09:30 A.M.

Home

Waning crescent

6/27 / 7:30 A.M.

45° / 80°E of S

10:00 A.M.

—

Waning crescent

6/28 / Cloudy

Work

Waning crescent

6/29 / 10:30 A.M.

60° / 70°E of S

11:30 A.M.

Work

New moon

6/30 / 10:30 A.M.

45° / 80°E of S

12:00 noon

Work

Waxing crescent

07/1 / 10:30 A.M.

55° / 45°W of S

12:30 P.M.

*Corrected by adding 0.5 hour for longitude position and adding 1.0 hour for Daylight Savings Time.

155

This experiment should be closely monitored, not just assigned and then ignored until the observation period is completed. Check the students’ progress each week, and make sure that they are using the proper observing and data-recording procedures. Students who complete this exercise usually become quite interested in observing the Moon, and whole families often get involved with these observations. One of our students reported that after several weeks of these “family” observations his 5-year-old brother stopped in the middle of a supermarket parking lot one afternoon and yelled “Look Mommy, there’s the Moon, there’s the Moon,” and nearly everyone in the parking lot stopped to look. Few people expect to find the Moon in the sky during the daytime hours. This experiment seems to be the kind of science project that successfully gets students and their families interested in observing the natural world around them.

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

Why does an observer on Earth always see the same side of the Moon’s surface? Ans: The Moon rotates on its axis with the same period as it revolves around Earth, so the same side always faces our planet.

During which phase of the Moon can a lunar eclipse occur? During which phase can a solar eclipse occur? Ans: Lunar eclipses occur at the full-moon phase; solar eclipses occur at the new-moon phase.

Make a drawing illustrating the positions of the Sun, the Moon, and Earth during a total solar eclipse. Show the shadows of both the Moon and Earth. (It is not necessary to make the drawing to scale.) Ans: See Figure 17.9 in the textbook.

The term terminator refers to the boundary line dividing day and night on the surface of a planet or moon. In this experiment, it is the line between the bright and dark side of the Moon. State the phase of the Moon when the terminator is the (a) sunrise line, (b) sunset line. Ans: (a) waxing phase; (b) waning phase.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What phase follows directly after the new phase of our Moon? Ans: Waxing crescent

How long does it take for the Moon to go from new phase to first-quarter phase? Ans: About 7 3/8 days

At what local solar time will the first-quarter moon be on an observer’s overhead meridian? Ans: 6:00 P.M.

156

What type of eclipse can be observed when part of Earth is in the shadow of the Moon? Ans: A solar eclipse will occur at this time for people within the shadow region.

The Sun and the Moon are both on an observer’s overhead meridian during what phase of the Moon? Ans: New moon

At what angle is the orbit of the Moon around Earth inclined with respect to the orbital plane of Earth around the Sun? Ans: Five degrees

157

Experiment 44

Hubble’s Law INTRODUCTION This is a math-oriented experiment in which students may have trouble converting units. This is to be expected, because students who dislike math often try to avoid problems that stretch their math skills. The experiment is fairly simple and easy to complete with a hand calculator. Present knowledge indicates that the value for Hubble’s constant is between 50 and 100 km/s/mpc. The distances in light-years to the galaxies in Figure 44.1 in the Laboratory Guide were calculated using a value of 50 km/s/mpc. The photographs in Figure 44.1 in the Laboratory Guide have been reproduced in many textbooks and periodicals with several different values for the distances to the galaxies. Don’t be surprised when you see several different values for the distances given, either in light-years, parsecs, or megaparsecs, because different values for the Hubble constant were used to obtain the distances. DATA TABLE 44.1 Distance in Light-Years

Distance in Megaparsecs

Recessional Velocity (km/s)

Virgo

7.80 × 107

23.9

1,200

50.2

Ursa Major*

1.00 × 109

306

15,000

49.0

Corona Borealis

1.40 × 109

429

22,000

51. 3

Boötes*

2.50 × 109

767

39,000

50. 8

Hydra

3.96 × 109

1220

61,000

50.0

Average value for Hubble’s constant

Hubble’s Constant (km/s/mpc)

50.3

* These are good data points but the photos are not included in this edition. You can supply them to the students if you wish to provide more points to graph.

159

70,000 Recessional velocity vs distance of star from Earth 60,000

50,000

40,000

30,000

20,000

10,000

rise 61,000 km s − 1,200 km s 59,800 km s = = run 1220 Mpc − 23.9 Mpc 1196 Mpc Hubble’ s constant = slope = 50.0 km s Mpc Slope =

500

1000

1500

Distance in megaparsecs

ANSWERS TO QUESTIONS IN THE MANUAL 1.

How does the slope of the curve compare with the average value of H calculated in Step 3? Ans: The slope should equal approximately 50 km/s/mpc which is Hubble’s constant.

Determine the value of Hubble’s constant in units of kilometers per year per kilometer. To do this, convert units of kilometers per second per megaparsec to kilometers per year per kilometer. Set the average value of Hubble’s constant plus units as recorded in Table of Conversion Factors equal to itself (see below). Multiply by the correct conversion factors, until the desired answer is obtained in the units you need. Show your work in the space below. 50

km/s/mpc =

km/s/mpc × (conversion factor)

See inside of manual’s back cover for appropriate conversion factors. Ans:

160

,400 s 50 km × 86day × s

365 days year

19 mpc km × 3.091×10 mpc 1

= 51.03 × 10 −12 year

The age of the universe (a rough estimate) can be obtained from the Hubble constant. The time elapsed since the Big Bang, which is the age of the universe, is the time of separation (distance d) of galaxies receding from one another with a velocity (v). From the equation for velocity v = d t we obtain

t =

d v

From Hubble’s law v = Hd , substituting, we get t =

d Hd

Canceling the d’s, we get t =

1 H

Calculate the value of t, in years, using the value for H determined in Question 2. Show your work. Here t is an estimate of the age of the universe. Ans:

= 19.6 × 109 years 5103 . × 10 −12 year This is about 20 billion years. t=

How does the calculated value of the age of the universe vary as the value of Hubble’s constant increases? Ans: The calculated age of the universe decreases as the value of Hubble’s constant increases.

How does the force of gravity between galaxies affect the recessional velocity of the galaxies? How would this affect the value of Hubble’s constant? The calculated age of the universe? Explain your answers. Ans: The effect of gravitational force on galaxies will decrease the recessional velocity of the galaxies. Hubble’s constant will be smaller because the constant is proportional to the velocity. This will give an increase in the calculated age of the universe.* * Current findings show, however, that this effect may be completely canceled by some unknown, even stronger repulsion force that appears to be causing the overall expansion rate of the universe to be accelerated, not slowing down at all.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

A cluster of galaxies has a recessional velocity of 100,000 km/s. Assume a value for Hubble’s constant equal to 50 km/s/mpc and calculate the distance to the cluster. v 100,000 km s Ans: = 2000 mpc d= = H 50,000 km s mpc

161

162

The calculated distances in light-years to the cluster of galaxies in Figure 44.1 are from present observed recessional velocities. Are the calculated distances the present distances to the galaxies? Ans: No. They are the distances from us at the time light left the cluster of galaxies.

Experiment 45

Measuring the Radius of the Observable Universe INTRODUCTION Many students do not understand that nearly all of the information that we have about the universe, other than for our own planet Earth, comes to us by the use of electrometric radiation. Our modern telescopes are not limited to optical and radio signals as they were in the past. Now we can detect radiation from the entire spectrum of EM signals and this gives us a much deeper understanding of the universe around us. Different types of electromagnetic radiation tell us about a great number of things. In this experiment we will concentrate on the ones that allow us to measure the distance to various parts of the universe. Modern telescopes and spectrometers make it possible to measure both the distance to far away galaxies and also their red shifts. This in turn has given us the ability to use these red shifts as a way to measure distance but this scale has had to be calibrated by using other methods of determining how far various celestial objects are from the Earth. Each procedure in this experiment deals with a specific use of electromagnetic radiation to provide data that can enable us to determine distances. The closest celestial objects, those within our own solar system can be measured using radar, that is radio waves. For more distant objects, that is the nearest stars in our Milky Way galaxy, we can use stellar parallax which requires the use of light signals. Then spectroscopic parallax comes into play using these same light waves but now analyzing them with a spectrometer to get even more information from them. Then pulsating stars lead us to information about their distance from us. For even longer distance measurements the Tully-Fisher relation used another spectroscopic clue, the line with of the hydrogen 21-cm radiation comes into play, and finally Hubble’s Law allows us to deal with distance measurements that extend to the very limits of our ability to detect electromagnetic radiation. The instructor should go over these various methods and show how standard candles, stars for which we believe that we know the absolute brightness from other measurements, provide the keys to using the inverse square law of distance for outgoing radiation to show us how far away things really are. Once the student has these basic concepts in mind, they can proceed to the various activities in this chapter and learn about making distance measurements throughout our universe. Please note that there are a couple of errors in the Laboratory Guide for this experiment. On page 297, Procedure 1, Question 1, the round trip time for radio waves should be 299.2 seconds. In Question 2, the unit on 0.300 should be A.U. not A.V. Finally the Electromagnetic Radiation Ladder, Figure 45.1, has several corrections that make it more useable.

163

PROCEDURE 1 1.

Calculate the distance to the planet Venus at its closest approach to Earth. The time for radio waves transmitted to Venus, at closest approach, and returned to Earth is 299.2 seconds to 4 significant figures. 299.2 d = vt = 3.00 × 105 km s s = 4.488 × 107 km = 4.50 × 107 km Ans: 2

FG H

IJ K

bor g = 186,000 mi sFGH 2992 .2 sIJK = 2.78 × 10 km 7

Using your answer to question one and knowing that Venus is 0.300 AU from Earth at its closest approach, calculate the length of one astronomical unit. Ans: Setup the following ratio and solve: 1 AU 0.300 AU = X 4.50 × 107 km

4.50 × 107 kg 1 AU . × 108 km = 150 0.300 AU

. × 108 km Therefore: 1 AU = 150 Electromagnetic radiation all wavelengths 7 × 108 ly

Hubble’s Law Step 6

To Observer

The Tully-Fisher relation Step 5

1.5 1067 lys 5××10 Cepheid variable stars Step 4 4 × 10 ly 4

Spectroscopic parallax Step 3 300 ly Stellar parallax Step 2 solar system only Radar ranging Step 1 Radio waves Earth Distance Ladder

Electromagnetic Radiation Distance Ladder to the Observable Limit of the Universe Figure 45.1 Steps used to measure successively more distant parts of the universe.

164

PROCEDURE 2 1.

The nearest star to Earth (excluding the Sun) is Proxima Centauri, which has a parallax of 0.772 arc seconds. Calculate the star’s distance from Earth. 1 d= = 129 . pc Ans: 0.772

Our Earth’s observatories cannot measure parallax smaller than 0.01 seconds of arc. Calculate the maximum distance, in parsecs, that we can measure to a star using stellar parallax. 1 d max = = 100 pc Ans: 0.010

The star Altair, in the constellation Aquila, has a parallax of 0.2 arc seconds. Calculate the distance, in light years, to the star. 1 3.26 light years Ans: d= = 5.0 pc = 16.3 light years 5.0 pc or 0.20 1 pc

FG H

b g

IJ K

The maximum base line for measuring parallax of stars is the major axis of Earth’s orbit around the sun. With our modern technology, suggest how astronomers can increase the base line. Ans: (1) Place the telescopes on Mars and use that planet’s semi-major axis for the base line. (2) Take measurements using space probes or satellites around other outer planets.

PROCEDURE 3 1.

Calculate the distance to a star with an apparent magnitude of 7 and an absolute magnitude of 2. Ans:

d = 10

m− M +5 5

= 10

7 − 2 +5 5

= 102 = 100 pc

Calculate the distance (in kilometers or miles) that light travels in one year. 365 days 24 hr 60 min 60 sec Ans: d = vt = 3.00 × 105 km s 1 year 1 year 1 day 1 hr 1 min 12 d = 9.46 × 10 km

gFGH

IJ FG KH

IJ K

Calculate the radius (in kilometers or miles) of the Hubble volume. Ans: If the age of the universe is taken to be 13.7 billion years 13.7 × 109 years :

jFGH 3.09 ×110pc km IJK γ = 4.19 × 10 pc babout 4 billion parsecsg e

γ = 9.46 × 1012 km 13.7 × 109 years

165

PROCEDURE 4 1.

On a clear night sky the galaxy Andromeda can be seen with unaided eyes. Was astronomer Hubble able to observe Cepheids in Andromeda? Justify your answer. Ans: Yes, since Andromeda is visible with unaided eyes, Hubble could easily detect Cepheids using the powerful 200-inch telescope. Andromeda is about 2.2 million ly from Earth.

Can the magnitude-distance equation be used to find the distance to pulsating Cepheids? Justify your answer. Ans: Yes, once the period of Cepheid is determined by observation, its period can be located on the period vs. absolute brightness graph and the absolute magnitude (M) can be found. Then if the apparent magnitude (m) is also measured, you can calculate the distance. d = 10

m − M +5 5

Name the two parameters that are directly measured by observing pulsating Cepheids. Ans: (1) period, and (2) apparent brightness can be directly measured for a Cepheid.

If the observations were made from the surface of the moon, could the distance to Cepheids be increased beyond 15 million parsecs? Justify your answer. Ans: No, a difference of 240 thousand miles closer to a Cepheid would not make any noticeable difference in the star’s apparent brightness and the Hubble telescope already gives us measurements that are not seen through Earth’s atmosphere.

PROCEDURE 5 1.

What is the significance of the Tully-Fisher relation? Ans: The Tully-Fisher relation gives astronomers another direct measurement tool for determining the absolute magnitude of a spiral galaxy using the width of H2 lines in the light spectrum that is independent of the radial motion measurements made using red-shifts.

What type of galaxy was observed by Tully and Fisher? Ans: A spiral galaxy.

What is the relation between galaxy brightness and the width of the emission lines? Ans: The wider the H2 line the brighter the galaxy.

How are the Tully-Fisher calculations for the distance to a galaxy different from those used for spectroscopic parallax? Ans: Since we are still using the apparent and absolute brightness measurements the calculations are the same but here we are using another independent variable to determine the absolute brightness.

166

PROCEDURE 6 1.

v 1 Show that H0 = . Where H0 = Hubble’s law, and by definition that d t distance d velocity v = . time t

Ans: 2.

bg bg

H0 =

v d t 1 = = d t d

If the age of the universe is really 13.7 billion years, calculate the value of the Hubble constant H0 . 1 1 Ans: H0 = = t 13.7 billion years 1 = 365 days 24 hr 60 min 60 sec 13.7 × 109 years 1 min 1 year 1 day 1 hr

FG H

H0 =

1 17

IJ FG KH

IJ K

= 2.31 × 10 −18 sec

4.32 × 10 sec To get this in the more familiar form we must introduce the units km Mpc . H0 = 2.31 × 10 −18 sec

F 3.09 × 10 km I = 71.4 km s Mpc GH 1 Mpc JK 19

Calculate the distance to a galaxy that has a recessional velocity of 50,000 km/sec using the value of H0 that you calculated above in Question 2. 50,000 km s v Ans: d= = = 700 Mpc H0 71.4 km s Mpc

ANSWERS TO QUESTIONS IN THE MANUAL 1.

Do you think that our universe is the only one that exists? Justify your answer. Ans: There is no data to suggest that our universe MUST be the only one that exists, and there is no data the PROVES that there is more than one universe like ours. This is much like the question about whether there is more than one galaxy in our universe of if the Milky Way is the only one. Until Hubble found experimental indications that there were stars too far away to be part of our own galaxy, people could believe whichever they chose. We are still waiting for some definitive data to answer this question so the student may express either opinion.

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What do you think exists beyond (a) our Horizon Volume (b) the radial limit of our universe? Current indications are that there is more of our universe outside of the Ans: (a) present Horizon Volume and this should look essentially the same as the portion of space that we find inside this boundary. (b) As with Question 1 above, this answer has no proven answer and there could be things quite different outside of the radial limit of our universe. It stands to reason that a student who believes that there is a possibility of more than one universe will have a different view of this question than a student who does not, but at this time almost any answer is acceptable since in this region “all things are possible”.

If we assume space to be infinite, what is the possibility of multiple universes where the laws of physics may be different than those of our universe? Ans: Since we have not detected a limit to space we must treat it as theoretically infinite, and since there are regions beyond the outer limits of our current observations we should probably admit the possibility of regions where the laws of physics are different. If such regions exist, however, it will be very difficult for us to understand what goes on there since our explanations of what we observe in space relies heavily on our belief that the principles of science that we see in action on and near our Earth, must also hold true for other regions of space as well.

How big is the universe? You calculated the radius of the Hubble Volume (the observable universe) in Procedure 3. Is it possible to calculate the radius of the entire (everything that is) universe? Justify your answer. (Hint: Consider the expansion rate.) Ans: Since space appears to be expanding and accelerating, the radius must be increasing. Our present data places no limits on this expansion and since it is space itself that is expanding, we cannot begin to calculate the location of an outer edge even if such an edge does exist. Because of this we cannot determine how big the universe is so no total radius can be calculated.

Ask yourself a question concerning the unknown beyond our universe. If you wish, answer the question. Ans: Students with good imaginations may have questions or views about what lies beyond our current limits of observation. Formulating and discussing such questions may be fun, but it must be remembered that any speculations about these regions of space are not backed up by any concrete data and must be treated as possibilities only.

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ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

In Procedure 6, Question 2, a conversion factor between km and Mpc was used. Derive this from the number of m in one parsec given in the table of conversion factors found on the last page of the Laboratory Guide. 1 km 1 pc = 3.09 × 1016 m = 3.09 × 1013 km Ans: 3 1 × 10 m Multiply both sides by 106 since there are 106 pc in one Mpc. 1 × 106 pc = 1 Mpc = 3.09 × 1013 × 106 km

FG H

IJ K

1 Mpc = 3.09 × 1019 km

If color is an indicator of the surface temperature of a star, would a reddish star in the sky have a higher or lower surface temperature than our yellow colored Sun? Ans: The reddish star would have a lower surface temperature than our Sun.

Would the reddish star in Question 2 give off more or less overall energy than our Sun? Ans: It depends on the size of the reddish star. If it had the same diameter as our Sun, it would give off less energy but if it was much, much larger (like some red giant stars really are) it could give off more overall energy.

From what you learned in Procedure 1, explain why there was a delay in the answers that our astronauts on the Moon sent back to NASA on Earth. Ans: It takes time for any EM radiation to travel from one distant point to another so the radio signals from NASA took some time to reach the Moon and the astronauts’ answers took some time to return from the Moon to Earth.

How long is the delay described in Question 4? The distance from Earth to the Moon is 2.4 × 105 km . 2.4 × 105 km d Ans: t= = = 0.80 seconds for the radio signal to travel 1 way, so the v 3.0 × 105 km s round trip would take 1.60 seconds.

What variables are plotted as the coordinates on an H-R diagram? Ans: The absolute brightness of a star is plotted as a function of the spectra class (or sometimes the color) of a star. Spectra class is closely related to the surface temperature of a star and thus its color.

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Experiment 46

Air Pressure INTRODUCTION No special introduction is provided for this experiment, but one or two demonstrations of air pressure can be performed easily. The Magdeburg hemispheres, if available, provide a good hands-on experience for students studying air pressure. If the hemispheres are not available, a metal 1-gallon can is the classic tool for demonstrating what atmospheric air pressure can do. Place approximately 1 in water in the can, boil the water, then screw the cap down tight. When the water cools, a partial vacuum forms, and the can is crushed by the air pressure.

ANSWERS TO QUESTIONS IN THE MANUAL 1.

If the atmosphere were condensed to the density of mercury, how high above sea level would the upper level of the liquid be? Use the data from this experiment to answer the question. Ans: The answer to this question, in inches and centimeters of mercury, is given in number 2 in the Procedure. The column of mercury is balanced by the weight of the atmosphere. (1 atm = 30 in of Hg = 76 cm of Hg)

What is the existing local air pressure in millibars? Show your work for obtaining the answer. _________ mb Ans:

Assume the present barometric pressure is 29.7 in of mercury. Refer to the introduction to find that 1 atm = 1013 mb = 30 in of mercury. 29.7 in Hg 1013 mb × = 1003 mb 1 30 in Hg

Is the air pressure in Question 2 considered to be a low or a high pressure when compared to normal atmospheric pressure? Explain. Ans: The answer depends on the barometric pressure of the neighboring surface area, but is somewhat lower than what is generally considered standard.

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What pressure units is a meteorologist using when he or she says the barometric pressure is 29.68 and falling? Ans: The barometric pressure reading of 29.68 has units of inches of mercury.

Using information gained from this experiment, explain why: a ship moving in a narrow canal may collide with the side of the canal. (a) (b) a person in a parked automobile along a freeway feels the car being “pulled” toward a semitrailer truck as it passes by. Ans: Both answers (a) and (b) are based on Bernoulli’s principle where the pressure on the side of a moving object will be lower on the side where the air is moving faster.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

State Bernoulli’s principle. Ans: In simple terms, Bernoulli’s principle states that where the velocity of a fluid is maximum, the pressure is minimum. Where the pressure is minimum, the velocity of the fluid is maximum.

In number 7 in the Procedure, what is the maximum possible pressure on the rubber base? Ans: The maximum pressure on the base will be a value equal to the atmospheric pressure in the room on that day.

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Experiment 47

Humidity INTRODUCTION Give a brief lecture stressing the difference between absolute and relative humidity, using examples. If a psychrometer is available, give a demonstration and determine the relative humidity of the air in the laboratory. Saturated water vapor tables are given in both SI units and British units. This allows the instructor to select the system he or she wishes to use in the laboratory. Although the introduction gives information and an example, some students will have trouble using the saturated water vapor tables to determine both absolute humidity and dew point, so it is worthwhile to explain how to use the tables, again in detail.

ANSWERS TO QUESTIONS IN THE MANUAL 1.

Explain the difference between absolute humidity and relative humidity. Ans: Absolute humidity is the amount of water vapor in a given volume of air and is usually measured in grains ft 3 or g m3 . One grain is 1/7000 lb. Relative humidity is the ratio of the absolute humidity to the maximum moisture capacity at a given temperature.

The air temperature in the laboratory is lowered 5°C. Assuming the absolute humidity remains constant, would the relative humidity increase or decrease? How much? Show your work. Ans: Assume the amount of moisture present to be 2.4 grains. 2.4 × 100 = 30% Relative humidity at 70° F = 8.4 2.4 Relative humidity at 65° F = × 100 = 35% 6.8

In this example the relative humidity increases by 5% but the actual moisture content of the air remains the same. If the relative humidity of the air in the laboratory is 25 percent at its present temperature, what is the absolute humidity? Ans: Obtain the air temperature in the laboratory. Refer to Table 46.1 and determine the moisture capacity at laboratory temperature. Absolute humidity = 0.25 × capacity. Example: If the room temperature is 68°F (20°C) the absolute humidity here is 17.3 g m3 ×.25 = 4.3 , or 7.9 grains ft 3 ×.25 = 2.0 grains . 173

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What is the absolute humidity of a saturated air sample at 21°C? Convert the Celsius temperature to degrees Fahrenheit, then use Table 46.1. Ans: 21°C = 70°F. The absolute humidity equals about 8.3 grains or 17.7 grams for a saturated air sample at this temperature.

The air temperature in your dormitory room is 80°F, and the relative humidity is 30 percent. A relative humidity of 50 percent is best for the human body. How can you increase the relative humidity of the air in your room without adding moisture to the air? Ans: Relative humidity can be increased by lowering the temperature of the air in the room. The absolute humidity here is about 25.5 grams* so 50% of this is 12.8 grams. This means the temperature would have to be lowered to 59°F which is quite chilly so this is not a very practical solution to the problem. * How do you arrive at 25.5 grams here? You can interpolate between reading in the table.

86°F 77°F 9° F

30.4 grams   subtract both 23.0 grams  7.4 grams

7.4 grams = 0.82 grams °F so 80°F requires 9° F 3°F increase above 70°F or 2.47 grams more water vapor. At 80°F the absolute humidity would be about 23.0 grams + 2.47 = 25.47 grams. Rounded to 3 figures this would give us the 25.5 grams reported above.

Divide number of grams by °F to get

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Experiment 48

Weather Maps (Part 1) INTRODUCTION This experiment acquaints students with the Daily Weather Map* issued by the National Weather Service and reinforces their knowledge of the concepts used to describe weather. For science laboratories not having access to maps from the National Weather Service, a Specimen Station Model and Weather Map Symbols are included in the experiment. These forms can be used as references for symbol notations and for determining the position of symbols around the weather station. The information on these forms is sufficient to answer all questions asked in the experiment. The instructor may want to change some of the city names listed in Data Table 48.1 to correspond to cities familiar to the students. The barometric pressures for the cities listed in Data Table 48.1 are shown on the weather map (Figure 48.3) in the Laboratory Guide. If other cities are used on this weather map, the barometric pressure values will have to be interpolated from the nearest isobar.

Bismark, ND Temperature, in degrees F Dew point, in degrees F Barometric pressure, in millibars

DATA TABLE 48.1 Oklahoma Montgomery, City, OK AL

Houston, TX

New Orleans, LA

1022.0

1020.0

1017.5

1016.5

1017.8

Wind direction*

WNW

Calm

Wind magnitude, in knots

13–17

3–7

8–12

*Daily Weather Maps can be obtained from the Superintendent of Documents, U.S. Government Printing Office, Washington, D.C. 20402.

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ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

List the changes in the temperature and the air pressure that occur as a warm front passes over a surface area. Ans: As a warm front passes over a surface area, the air temperature rises and the air pressure generally drops.

List the changes in the temperature and the air pressure that occur as a cold front passes over a surface area. Ans: As a cold front passes over a surface area, the air temperature drops and the air pressure generally rises.

The Surface Weather map presents data and analysis for what time of day? Ans: The Surface Weather Map presents data and analysis for 7 A.M. EST.

Define an isobar. How are they represented on the weather map? Ans: An isobar is a line of equal pressure. Isobars are represented as solid lines on the weather map.

Define an isotherm. How are they represented on the weather map? Ans: An isotherm is a line of equal temperature. Isotherms are represented as short dashed lines on the weather map.

How is the amount of precipitation indicated on the weather map? Ans: The amount of precipitation is indicated on the weather map as a number in inches during the past 6 hours. The number is located at the six-o’clock position of the weather station.

What is the lowest pressure indicated by an isobar on Fig. 48.3? Ans: 96 mb over Iowa

What is the highest pressure indicated by an isobar on Fig. 48.3? Ans: 1028 mb over Saskatchewan

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

How is the direction and magnitude of a wind indicated on the weather map? Ans: A straight line indicates direction. The number of flags on the line indicates magnitude.

Distinguish between an occluded front and a stationary front. Ans: An occluded front is the boundary between two air masses where a fast-moving cold front is overtaking a warm front. A stationary front is the boundary between a warm front and cold front. The opposing fronts balance each other so no movement occurs.

How is a thunderstorm indicated on the weather map? Ans: The following symbol represents a thunderstorm.

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Experiment 49

Weather Maps (Part 2) INTRODUCTION A brief lecture should review Experiment 48 Emphasize the concepts of high- and low-pressure cells. Define the bar and give the conversion factors needed to change millibars to inches of mercury. The answer to Question 2 is found on the back of the Sunday issue of the Daily Weather Maps, Section 12, or on the weather map explanation sheet put out by the Environmental Data Service. These explanation sheets are for sale by the Superintendent of Documents, U.S. Government Printing Office, Washington, D.C. 20402. They are sold in 50-copy lots at a reasonable price. You should have a map of the United States in the laboratory or be prepared to answer questions concerning location of cities in states that have more than one city listed. You could make a copy of the map for number 5 in the procedure, write in any other cities that you want the students to examine during this experiment, and post the map on the bulletin board or place a copy on each lab table.

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

How is the unit of pressure, the bar, defined? (See Experiment 46.) Ans: The unit of pressure, the bar, is defined as 106 dynes/cm2. One bar equals 1000 millibars (mb). One atmosphere equals 1013.2 mb.

What is the normal range of atmospheric pressure (in millibars) at Earth’s surface? Ans: The usual range of atmospheric pressure (in millibars) at Earth’s surface is approximately 950 to 1050 mb.

What is the difference in pressure indicated by two adjacent isobars on a weather map? Ans: The difference in pressure indicated by two adjacent isobars is 4 mb.

What is the direction of the air circulation around a low-pressure cell in the northern hemisphere? Ans: The direction of air flow around a low-pressure cell in the northern hemisphere is counterclockwise.

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What kind of weather is usually associated with a low-pressure cell in North America? Why? Ans: Bad weather is usually associated with a low-pressure cell. Cloud cover and precipitation can usually be expected in and around low-pressure cells.

What is the direction of the air circulation around a high-pressure cell in the northern hemisphere? Ans: The direction of air flow around a high-pressure cell in the northern hemisphere is clockwise.

What kind of weather is usually associated with a high-pressure cell? Why? Ans: Good weather is usually associated with a high-pressure cell because moisture laiden air is kept from entering the area.

A pressure of 1012 mb is indicated on a weather map. Is this pressure high or low? Explain. Ans: A pressure reading of 1012 mb may indicate a high- or low-pressure cell. The reading depends on the value of the pressure for the surrounding area but the reading is a little lower than the value generally associated with normal atmosphere pressure at sea level.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

Distinguish between a high- and a low-pressure cell. Ans: In a high-pressure cell the air is heavy and descending. The air at ground level is leaving the cell to move to neighboring areas, and the air circulation is clockwise. In a low-pressure cell the air is lighter and rising. The air from neighboring areas is entering the cell at ground level, and the air circulation is counterclockwise around the center of the cell.

At Earth’s surface what is the direction of the air flow between a high- and low-pressure cell? What is the direction aloft? Ans: At the surface the air flow is from the high- to the low-pressure cell. Aloft the air flow is from the low- to the high-pressure cell.

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Experiment 50

Topographic Maps INTRODUCTION The U.S. Geological Survey maps called quadrangles are issued in three common sizes. They have scales as follows: 1:250,000 covers an area of 2° longitude and 1° latitude; 1:62,500 covers an area of 15 min by 15 min; 1:24,000 covers an area of 7 min 30 sec by 7 min 30 sec. A brief lecture on topographic maps is a must. Don’t assume students know the definitions of latitude and longitude. Define, explain, and give examples of each by referring to the values given on the topographic maps. Explain the scale of the map and show how to calculate the area of the map in units of square miles (1 min of arc equals 1 nautical mile or 1.15 land miles). Indicate to students that the topographic map portrays both horizontal and vertical distances. Explain how to obtain each. The answers to the questions for this experiment depend on the quadrangle map selected by the instructor.

ANSWERS TO QUESTIONS IN THE LAB MANUAL 1.

What are quadrangle maps and what governmental agency produces them? Ans: Quadrangle maps show both the horizontal and vertical layout of the land. U.S. Geological survey.

What can be said about the terrain shown on a topographic map where the contour lines are very close together? Ans: The elevation of the land is changing rapidly.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What is the latitude and longitude of the point at the lower right-hand corner of the topographic map? The point at the upper left-hand corner? Ans: The answer is given at the corners indicated in the question and is written directly on each map.

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If you have a topographic map of the local area, what is the latitude and longitude of your school? Ans: Obtain the answer from the x and y coordinates of the map.

State the scale of your topographic map in units of inches per mile; that is, one inch equals how many miles? Ans: The answer depends on the scale of the map the students are using and can be found on the map key printed on each map.

What is the distance (in feet) indicated by adjacent contour lines? How is a difference of 100 ft indicated on the map? Ans: The distance between contour lines is usually 20 ft. The 100-ft differences are indicated by a heavier line.

Where is the highest point on the topographic map? Determine the average slope due north and due south of the highest point. Give your answer in feet per mile. Slope equals the change in elevation divided by the change in horizontal distance. Ans: The answer depends on the map the students are using.

What is the difference in elevation, measured in feet, between the highest and lowest points on the map? Ans: The answer depends on the map the students are using.

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Experiment 51

Minerals INTRODUCTION Many students find the study of rocks and minerals fascinating, but most have a difficult time identifying them. Arouse students’ interest with a brief lecture on gems such as diamonds, emeralds, rubies, and sapphires. State their chemical composition and explain their crystalline structure. If you can, tell something of their origin and where they are found. Also, using an ultraviolet light to show fluorescent minerals appeals to students. You also might give a brief lecture on the physical properties of minerals and how they are identified.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What is the composition of minerals? Ans: Minerals are composed of elements.

Discuss the crystalline structure of minerals. Ans: A crystalline substance has a definite internal structure such that the atoms are in specific proportions and are arranged in an orderly geometric pattern. Under ideal conditions of crystal growth, perfect crystal faces will form. Although the size of a crystal may vary, the angle between faces of a specific mineral will be the same. Substances that when crystallizing become solid without a definite internal structure are called amorphous.

Distinguish between cleavage and crystal faces. Ans: Natural crystal faces form and produce perfect geometric patterns when a substance is crystallized. Cleavage develops from planes of weakness within the crystal structure. The crystal will break along the weakness planes.

Define and give an example of tenacity. Ans: Tenacity refers to the ability of a mineral to hold together. Quartz, a very brittle mineral, has low tenacity. Biotite, a very elastic mineral, has high tenacity.

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182

Distinguish between fluorescence and phosphorescence. Ans: Fluorescence is the immediate emission of radiation, such as visible light, produced by the absorption of other radiation of a higher frequency. Phosphorescence is the persistent emission of light following exposure to and removal of incident radiation from the sample.

Experiment 52

Rocks INTRODUCTION Using some common specimens of each rock classification, give an introduction to their physical characteristics, composition, and how they are identified. Emphasize the fact that nearly 75 percent of Earth’s crust consists of only two elements, oxygen and silicon, and almost 99 percent is composed of these two plus six others. Listed in order of abundance, they are aluminum, iron, calcium, sodium, potassium, and magnesium. Define and explain silicates. Minerals that contain oxygen and silicon plus one or more of the other six elements are known as silicates. Most rocks are aggregates of minerals. Thus, to identify and classify rocks, we need to know (1) what minerals make up the rock’s composition, (2) the texture of the rock—that is, grain size, and (3) how the grains are arranged.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

How is a rock defined? Ans: A rock is any naturally occurring, solid mineral mass that makes up part of Earth’s lithosphere.

How are rocks classified? Ans: Rocks are classified according to how they formed or originated.

State the three chief classifications of rocks. Ans: Rocks are classified as igneous, sedimentary, or metamorphic.

What are the two most abundant elements found in Earth’s crust? Ans: The two most abundant elements are oxygen and silicon.

What are silicates? Ans: Silicates are rock-forming minerals composed of one or more of the metals plus silicon and oxygen.

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Experiment 53

Rock-Forming Minerals INTRODUCTION Briefly review the introduction to Experiments 51 and 52, adding basic information on some of the chemical groups that comprise the rock-forming minerals. Begin with the feldspars, since they are the most abundant. Orthoclase feldspar, KAlSi3O8, is usually pink to flesh colored and is the dominant mineral in granite. A feldspar specimen should be passed among the students during the lecture. Following this, give information on the silicates (quartz and olivine), the micas (biotite and muscovite), the oxides (hematite and magnetite), the sulfides (pyrite and galena), the sulfates (gypsum and anhydrite), and the carbonates (calcite and dolomite).

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What is the composition of silicates? Ans: Silicates are composed of one or more of the metals plus silicon and oxygen. Some rocks contain several silicate groups.

Name the chemical group most abundant in the rocks of Earth’s crust. Ans: The feldspars form the most abundant group of rock forming minerals.

What is the chemical composition of quartz? Ans: Quartz is silicon dioxide (SiO2).

Distinguish between the micas, biotite, and muscovite. Ans: Biotite and muscovite are both members of the family of minerals known as micas. Biotite is a dark-colored mica, brown, black, or sometimes green, containing iron and magnesium. Muscovite is a light, almost colorless mica containing potassium and aluminum. Mica has perfect cleavage in one direction.

What is the common name for pyrite (FeS2)? Ans: The common name for pyrite is “fool’s gold.”

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Experiment 54

Igneous Rocks and Crystallization INTRODUCTION Because a rock is an aggregate of minerals and a mineral is any naturally occurring, inorganic, crystalline substance, an experiment showing crystal growth provides a better understanding of the crystalline structure of minerals. There will be time to show a film on crystal growth if one is available. Films can be obtained on loan from the U.S. Geological Survey. See the Instructor’s Guide for the textbook for the address. Caution students not to handle the thymol with their bare hands. Although thymol is not poisonous, it may irritate the eyes and skin.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

How does the rate of cooling affect grain size? Ans: Slow cooling produces large-sized grains. Fast cooling produces small-sized grains.

Define texture. Ans: Texture refers to grain size—that is, glassy, fine, medium, or coarse grain, or mixed.

What does the texture of a rock indicate about its cooling process? Ans: The texture is a good indicator of how the magma cooled. For example, a porphyritic texture indicates both a slow- and a fast-cooling process took place at different times as the magma cooled.

What are the two criteria used in Table 54.1 to identify and classify igneous rocks? Ans: Texture and mineral composition

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Experiment 55

Sedimentary Rocks INTRODUCTION You might give a brief lecture on the composition of sedimentary rocks emphasizing the four major constituents—quartz, calcite, clay, and rock fragments—and show a short film on the major processes involved in the genesis of sedimentary rocks. Point out that sedimentary rocks supply geologists with the most accurate information, because they are the most abundant rock and the processes that produce them are very accessible, in contrast to igneous and metamorphic rocks.

ADDITIONAL OR ALTERNATIVE QUESTIONS 1.

What are the four materials most often involved in the making of sedimentary rocks? Ans: Quartz, clay, rock fragments, and calcite

Name the three groups of sedimentary rocks Ans: Clastic, biochemical or chemical, and organic

Why are sedimentary rocks and the processes that produce them more accessible than igneous and metamorphic rocks? Ans: Sedimentary rocks are formed at or near Earth’s surface. Most igneous and all metamorphic rocks are formed deep inside Earth.

What is the meaning of porosity? Ans: Porosity refers to how permeable to liquids a substance’s composition is, that is, how much water could be held in that type of rock sample.

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Appendix

Integrated Equipment and Supply Lists INTRODUCTION Laboratory experiments for the Laboratory Guide have been designed with the needs of the prospective elementary-education teacher in mind. Three principles determined the choice of experiments: 1.

Present basic scientific principles assigned and discussed in the lecture class as simply as possible.

Keep the cost of the equipment and supplies to a minimum.

Select experiments that can be duplicated by students later when they are themselves teaching.

In this equipment list for all 55 experiments, only five pieces of equipment cost more than $100. The most expensive experiments deal with radioactivity. The air track is expensive, but the wooden plank is a good substitute. The balance, inclined plane, and celestial sphere (globe) each cost less than $200. To assist the instructor in planning the course, the equipment and supply lists for the experiments are listed by scientific discipline. This arrangement necessarily includes a few duplications. Note also that each list may not contain every small supply item.

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PHYSICS EQUIPMENT Balance, Triple beam or electronic Boyle’s law apparatus Bunsen burner and gas lighter Calorimeter Centripetal-force apparatus Clamps (assorted) Color filters (plastic transmission, assorted colors) Compass (magnetic) Diffraction grating and holder Dip needle Electroscope Fisher light and optics set Hall’s carriage High-intensity light Inclined plane Lamp (sodium vapor or mercury arc) Lens (convex) Linear air track or wooden plank Magnets (bar) Masses (hooked and slotted gram sets)

Meter sticks (one- and two-meter) Micrometers or calipers Milliammeter (0–500 mA, dc) Mirror (concave) Moment-of-force apparatus Pendulum (simple) Pulleys (assorted) Pulley (precision ball bearing) Radiation detector with power supply Resonance apparatus Rheostat (40-ohm) Rutherford scattering box Slit with attached scale Spring (k = 20,000 – 30,000 dynes/cm) SPST switch Steam generator Timing device Tuning forks Vacuum base Voltmeter (0–10 V, dc)

SUPPLIES Balloons (toy) Beakers (assorted) Candles Cardboard screen Cloth, wool (or fur), silk Coiled spring Color chips (from paint store) Deflagrating spoon Dry cells (1.5 V or 6.0 V) Fuse holder Glass marker Keuffel and Esser wave-pattern transparencies Lead shielding Lens filter (colored) Lens holder Masking tape Metal can (large coffee) Metal pellets (aluminum or copper)

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Paint brushes (small) Radiation sources Resistor (40-ohm, 5-watt) Rods (hard rubber and glass) Rubber hammer Rubber tubing Slinky Tempera paint Stopper (cork) String Support rods and holders Trays (large, water) Tripod base for glass beaker Weight hangers for slotted weights (masses) Wing tip for Bunsen burner Wire (electrical hook-up) Wooden blocks (small) Wooden splints

CHEMISTRY EQUIPMENT Bunsen burner and gas lighter Burettes Calorimeter Clamps (assorted) Combustion tube Coulometer Cylinders (graduated 100 mL) Evaporating dishes Erlenmeyer flasks

Milliammeter (0–500 mA, dc) Molecular-models kits Pneumatic trough Pycnometer Ring stand Stirring motor Support rods and holders Trays (large, water)

SUPPLIES Chemicals (assorted) Filter paper Funnel Glass marbles Glass wool Glassware; Beakers (Pyrex), Bottles, Stirring rods, Test tubes

Metal pellets Rubber tubing Stoppers (assorted) Thermometers Wire (steel, picture) Wood splints

ASTRONOMY EQUIPMENT Celestial sphere

SUPPLIES Rulers Masking tape

Protractors

METEOROLOGY EQUIPMENT Calorimeter Meter stick

Tray (large, white, enamel) Vacuum base

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SUPPLIES Balloons (toy) Beakers (Pyrex) Cardboard index cards Clamps (pinch) Glass plates (small) Ping-pong balls Plastic bottles (250 mL) Plastic spoon (small)

Rubber bands Rubber tubing Soda straws (plastic) Straight pins Thermometer Tumbler (small, glass) Weather maps

GEOLOGY EQUIPMENT Compass (magnetic) Magnifying glass

Millimeter rule Ultraviolet lamp

SUPPLIES Acid (hydrochloric, dilute) File (steel) Forceps Glass jar (small) Glass slide or evaporation dish Knife

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Liquid dropper Topographic quadrangles Rock and mineral specimens Sand (coarse) Sand mixture Solution of several salts