Physics Standard level Paper 2 Monday 3 May 2021 (afternoon) 1 hour 15 minutes Answer ] with • • plant + " ° " + it
1. Two players are playing table tennis. Player A hits the ball at a height of 0.24 m above the edge of the table, measured from the top of the table to the bottom of the ball. The initial speed of the ball is 12.0 m s-1 horizontally. Assume that air resistance is negligible. (a) Show that the time taken for the ball to reach the surface of the table is about 0.2 s. (b) Sketch, on the axes, a graph showing the variation with time of the vertical component of velocity vv of the ball until it reaches the table surface. Take g to be +10ms-2. [2] 0 Consider vertical motion between point 0 & P s= UH ) 1- Kath 024=0 ( t) % (9.8×1-2) p t2= ' 2¥82 0.2sec =t § Vg=ug + Agt Vg -104 ) vy=o gt Vy= gt ,tank 10 Vg = 1011-7
ball
speed of
a peak height of
m
mass of the ball
(Question 1 continued) (c) The net is stretched across the middle and the net has a height of 15.0 cm. Show that the ball will go over the net of the table. The table has a length of 2.74 m. (d) The
bounces and then reaches
0.18 m above the table with a horizontal
10.5
s-1. The
is 2.7 g. (i) Determine the kinetic energy of the ball immediately after the bounce. (ii) Player B intercepts the ball when it is at its peak height. Player B holds a paddle (racket) stationary and vertical. The ball is in contact with the paddle for 0.010 s. Assume the collision is elastic. Calculate the average force exerted by the ball on the paddle. State your answer to an appropriate number of significant figures. [3] M = en , } @ ' 24-015 ) 7- o.ie mt 315cm ¥4 V L sa = Unt + % art ← 2- 74 Mt → ¥ 4 = (2) (t) t= 2 24 ← → Sg = oxt b. ✗ 9-8×1-2 ↳ Net Middle of Sg = ¥ ✗ 9.0×229*-3-474=6-38 Mt = v06 Mt the table ↳ much over tore height of the net Abbey conservation of energy between point M & N mgh + t.mg?-o-tmvi2;7-opo✗ 10 ✗ • 24 + { Fg 6123 = { MVP • 00640+1.944 = 1- moi 1- 955 = Iz MII ANS F. At = M (DD = MC Vf Vi ) d (c) f = } ˢoog4°÷ ◦ ,8"=%ooE ¥ % = 5 67N I 5- 7 N
2. A planet is in a circular orbit around a star. The speed of the planet is constant. The following data are given: Mass of planet = 8.0 × 1024 kg Mass of star = 3.2 × 1030 kg Distance from the star to the planet R = 4.4 × 1010 m. (a) Explain why a centripetal force is needed for the planet to be in a circular orbit. 7 24 30 ☒ Aaaaa qq.gg * %☆@•t I % ANS 2 (a) ^ Fe for circular motion LfgGMs Mp mob ( bttne STAR 82 = -5 Centre ] → circular motion involves charge in velocity → centripetal acceleration is perpendicular to the torrential velocity
IB PHYSICS TUTORS IN USA
IB PHYSICS TUTORS IN INDIA
IB PHYSICS TUTORS IN UK
IB PHYSICS TUTORS IN UAE
IB TUTORS IN CANADA
IB TUTORS IN GERMANY
IB TUTORS IN AUSTRALIA
IB TUTORS IN SWITZERLAND
IB TUTORS IN POLAND
IB TUTORS IN SPAIN
IB TUTORS IN CHINA
IB TUTORS IN ECUADOR
IB TUTORS IN HONGKONG
IB TUTORS IN TURKEY
IB TUTORS IN MEXICO
IB TUTORS IN INDONESIA
IB TUTORS IN FINLAND
IB TUTORS IN NORWAY
IB TUTORS IN SWEDEN
IB TUTORS IN DENMARK
IB TUTORS IN RUSSIA
IB TUTORS IN EGYPT
IB TUTORS IN FRANCE
IB TUTORS IN PORTUGAL
IB TUTORS IN ARGENTINA
IB TUTORS IN ABU DHABI
IB TUTORS IN OMAN
IB TUTORS IN QATAR
IB TUTORS IN JAPAN
IB TUTORS IN PERU
IB TUTORS IN ITALY
IB TUTORS IN MALAYSIA
(b) Calculate the value of the centripetal force
f- = 9 Mi Mz f= 667 × 15 " " ✗ 8.0×10>4×3-2×1030 (4.4×1010) * = £67 ✗ 8.0×3.2 ✗ 10-111-241-30 4.4×4.4 ✗ 1020 = (6.67×8-0×3.2) ✗ 10-11-124 -130-20 4-4 × 4-4 = 8.81 ✗ 1033 Newton
(c) A spacecraft is to be launched from the surface of the planet to escape from the star system.The radius of the planet is (i) Show that the gravitational potential due to the planet and the star at the surface of the planet is about J kg-1. (ii) Estimate the escape speed of the spacecraft from the planet–star system. . 9 'll ✗ 103km ' 5×109 tlotential U = Ggb= 6.67×1534×8.0×1024 ° 91×103 ×103 = 5.8×107 %g Us = G- = 6-67 ✗ 10-34×3.2×1030 4^4×1610 = -4-85×109%9 Total potential V= Up -1 Us = -5 × 109 Jlkg F- TOTAL > 0 , F- K > / Ep / = mug m÷ > Hug V2 > zvg > 2×4-9×109 U > 1-2×4.9×109 = 9.91×104 m 5 '
New York Physics Tutor
North York TutorsMarkham TutorsVaughan
TutorsMississauga TutorsBrampton
TutorsBradford West Gwillimbury TutorsOshawa
TutorsHamilton TutorsGuelph TutorsNiagara Falls TutorsNorth Tonawanda TutorsTonawanda
TutorsKitchener TutorsWaterloo TutorsBrantford
TutorsWilliamsville TutorsBuffalo
TutorsCheektowaga TutorsAkron TutorsWest Seneca Tutors
Toronto Physics TUTOR Toronto ACT TutorsToronto Algebra TutorsToronto Biology TutorsToronto Calculus TutorsToronto Chemistry TutorsToronto English TutorsToronto
French TutorsToronto Geometry TutorsToronto
GRE TutorsToronto ISEE TutorsToronto LSAT
TutorsToronto Mandarin Chinese TutorsToronto
Math TutorsToronto MCAT TutorsToronto Physics
TutorsToronto Precalculus TutorsToronto Reading
TutorsToronto SAT TutorsToronto Spanish TutorsToronto SSAT TutorsToronto Statistics
TutorsToronto Writing Tutors
PHYSICS TUTOR IB SL CANADA
3. A mass of 1.0 kg of water is brought to its boiling point of 100 °C using an electric heater of power 1.6kW. (a) (i) The molar mass of water is 18 g mol-1. Estimate the average speed of the water molecules in the vapor produced. Assume the vapor behaves as an ideal gas
KT mob a' 3¥ = fÑa) mass __ ¥a = [3×1.38×-10%3] 3) ✗ 6.02×10231 18×10-3 = J 3×1.38×373×6.02 18 ✗ ( 4- 720m51
(ii) State one assumption of the kinetic model of an ideal gas. (b) A mass of 0.86 kg of water remains after it has boiled for 200 s. (i) Estimate the specific latent heat of vaporization of water. State an appropriate unit for your answer. Explain why the temperature of water remains at 100 °C during this time particle can be considered points ( ullth out dimension) NO Intermolecular forces volume of the particle is negligible to the volume of the gas water vapor is ed = 1- 0.86 P = ¥ w =P Lt > M L = 1.6×103×200 = 014kg L = 1^6×103×200 = 2-3×106 %g014 ( Ii ) All the em ogg added to use to break the bond
The heater is removed and a mass of 0.30 kg of pasta at -10 °C is added to the boiling water. Determine the equilibrium temperature of the pasta and water after the pasta is added. Other heat transfers are negligible. Specific heat capacity of pasta = 1.8 kJ kg-1 K-1 Specific heat capacity of water = 4.2 kJ kg-1 K-1 C. Heat lost by the water = Heat gained by the pasta M water swale ✓ ( 100 e.) = Mpasta Spas ta ( T C- 107 ? 086 × 42 × (100-7) = 0-30 ✗ 1.8 ✗ (7-110) 3 6 12 ✗ 100 3.612T = O ' 54T + 054 ✗ 10 361.2 5-4 = (3-6 121-054) T T= 355.8 4. 152 ʰ 86%
(ii) Calculate the power transferred by the heater when both switches are closed. P=V¥= I -611103W 1.6 × 103 ¥2 11 R= 220×220 1^6×103 = ¥¥ ! = 12¥ I 30 ohhh R Tmr nÉp' = ¥ ,-52 ( %) zaovoltLmf ↳ a = 2× 1.6× 103 w ↳ ← 220 Volt = 3.2×103 W' =3 -2kW = 3200 Watt
4. A planet orbits at a distance d from a star. The power emitted by the star is P. The total surface area of the planet is A. (a) (i) Explain why the power incident on the planet is P ×A. → PLANET ORBat Ffr surface Area of thanet A- = 47T R& fSTAR / d → )4- = TTR ' ✓ * A P= PowerPower radiated by the emitted by the = SURFACE star on unit area at Star AREA distanced is = -P of 49 DI PLANET Power incident on the planet __ ¥d ) (F)
(ii) The albedo of the planet is αp . The equilibrium surface temperature of the planet is T. Derive the expression where e is the emissivity of the planet. Permitted = e 6 AT 4 albedo = Total scattered power total incident power = Lpfrom previous tart P scattered = @ p ) ( Pine ident ? P incident =Pscattered + Permitted ¥ ) (F) µ P incident = Xp Pine ident + e • At 4 P incident ( a- Xp ) = e. 6 AT 4 74 = P incident ( a- Xp ) e. 6 A P A- ( e xp ) _ Proved 1- = 4 Fadi ETA
Brampton Tutors
Calgary Tutors
Edmonton Tutors
Halifax Tutors
Hamilton Tutors
Mississauga Tutors
Montreal Tutors
Oshawa Tutors
Ottawa Tutors
Quebec City Tutors
Toronto Tutors
Vancouver Tutors
Victoria Tutors
Waterloo Tutors Windsor Tutors
Winnipeg Tutors
Home Tutors in Kolkata, Physics Tutors in SaltLake,Kolkata, Physics Tutors in Kolkata, Online Physics Tutors in Kolkata, Physics Tutors in Bidhan Nagar, Kolkata,Physics Tutors in Jadavpur,Kolkata,Online Physics Tutors in Delhi, Physics Tutors in Teghoria,Kolkata,Physics Tutors in Baguiati,Kolkata,Home Tutors in Mumbai, Physics Tutors in Ballygunge,Kolkata,Online physics Tutors in Jaipur, Physics Tutors in Tollygunge,Kolkata,Physics Tutors in Mumbai, Online Physics Tutors in Mumbai, Physics Tutors in South Kolkata, Home Tutors in Delhi, Physics Tutors in Delhi, Online Physics Tutors in Patna, Home Tutors in New Delhi, Physics Tutors in New Delhi, Home Tutors in Patna, Physics Tutors in Patna, Physics Tutors in Jaipur, Home Tutors in Jaipur, Physics Tutors in Gurgaon,
(b) On average, the Moon is the same distance from the Sun as the Earth. The Moon can be assumed to have an emissivity e = 1 and an albedo αM = 0.13. The solar constant is 1.36 × 103 W m-2. Calculate the surface temperature of the Moon.
✗ $3 P= 4 g÷× 4- Xp ) Get = 1.36 × 103 × (0-87) 4 × 5.67 × 10-8 T= 268-75 K T = 270K
Bqz V23 8- Th > 34 + , He 't 90 2 06 20 PbTl → • z 1a81 to measure Radioactive decay for either
Gurgaon is a happening city with a booming infrastructure with rapid urbanization, economic prosperity, and intellectual growth. It boasts of being an abode of some top progressive educational institutes of the country catering to modern facilitation approaches.
If you are searching for IB online tutors in Gurgaon at some of the key locations like Sector 56, DLF City Phase 3, DLF Phase 5 Golf Course Road, Sushant Lok-2 and Acharyapuri, Sec 56, Gurgaon HO, you have landed up on the right website. Sev7n is a Gurgaon-based IB tuition class specialising in IB Maths, Math IA, BM IA, Extended Essay, EE, Physics IA, Chemistry IA, Biology IA, Bio IA, and IB TOK.
Our experienced IB Tutors in Gurgaon provide personalized home tuitions through IB online tutoring services.
IB Tutors in Gurgaon near me
We have identified lots of searches for IB tutors by locality like Sector 14, Sushant Lok 1, Sector 56, Sector 52, Sector 7, Palam Vihar, Sector 24, Sector 49, Sector 17, Sector 15, Sector 23, Sector 57, DLF Phase 1, Sector 4, Sector 54, Gurgaon Industrial Estate, Sector 11, Sector 31, Sector 40, Sector 45, Sector 46, Galleria DLF Phase 4, DLF Phase IV, New Railway Road, Sector 10, Sector 48, Sector 50, Sector 51, Sector 13, Dlf City, DLF Phase 2, Sector 5, Sector 55, Sector 12, South City II, New Palam Vihar, Sector 10A,
High temp = more KE , more speed ↳ energy required to overcome electrostatic repulsion
energy Is released when binding enemy bet nucleon increases 144 q , V > 35 Kyle + Ba 1- Q I 1↓ 89× 8 6. trev 144 × 8 2. Her 235×76 MeV Q = ( (23530-6) ( 8 (8-6) (144 × 8-2) ] = 160 Mev
6. On a guitar, the strings played vibrate between two fixed points. The frequency of vibration is modified by changing the string length using a finger. The different strings have different wave speeds. When a string is plucked, a standing wave forms between the bridge and the finger. (a) Outline how a standing wave is produced on the string. [2] b) The string is displaced 0.4 cm at point P to sound the guitar. Point P on the string vibrates with simple harmonic motion (shm) in its first harmonic with a frequency of 195 Hz. The sounding length of the string is 62 cm. (i) Show that the speed of the wave on the string is about 240 m s-1. Travelling wave moves along the length of the string and reflects at fired end And because of the superposition of Incident hid reflected wave , the standing wave is produced N H / A / Amplitude = 046M , f , = 195 Hz , L = 062Mt ← L = % L = ✗ = 2 ( L) = 2 ( 0627 = 1.24 Mt D= f- ✗ = (1-95) (1-24)=242 Mlseo
The team of expert tutors who form an integral part of the IB tuition in Gurgaon teach IB courses online. They cater to the individual needs of every student studying the IB Diploma Programme. IB Diploma Programme is very rigorous, comprehensive, unique, and advanced in its own way of infusing the critical thinking power in students. The IB Diploma Programme Course goes beyond the basic understanding of theories and concepts. It inculcates thinking skills, reflection, open-mindedness, empathy, balance, and risk-taking attitude in the learners through its profound learner profile. Redundant to mention that all our team of tutors and surely IB tuition in Gurgaon tutors are completely well versed with the meaning of the International Baccalaureate curriculum, its aims and objectives, assessment criteria, and the way how the IB Learner Profile is weaved through all of these aspects of teaching and learning.
Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position.
Iii ) For SHM A ✗ x
7. Conservation of energy and conservation of momentum are two examples of conservation laws. (a) Outline the significance of conservation laws for physics. (b) When a pi meson π- (du ) and a proton (uud) collide, a possible outcome is a sigma baryon Σ0 (uds) and a kaon meson Κ0 (ds ). Apply three conservation laws to show that this interaction is possible. → Both express to in cities of nature → feat = d÷ , If fexternae = 0 , b= contrail → ( TE ) = ( KE) + ( P E) = Constant , can aptly to two variable position and fend out unknown quantity •• qq0 DT + UUD → Ud Stds Using idea that particle and antiparticle has opposite number 2d tñtw + U →2d -1k -1s -15 u=U Conservation : charge ( Yes ) 0=0 baryon number Yes 1-3=1-3 strangeness → Yes
KUMAR PHYSICS CLASSES E 281 BASEMENT M BLOCK MAIN ROAD GREATER KAILASH 2 NEW DELHI +91- 9958461445 www.kumarphysicsclasses.com www.kumarneetphysicsclasses.com IB PHYSICS SL PAPER-2 3 MAY-2021 SOLUTION WITH EXPLANATION
