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MBM AJ 700 Air-Feed Paper Jogger February 6th, 2012 |
Author: babylover
A In-Depth Look at the MBM AJ 700 Air-Feed Paper Jogger Does your workplace need a durable, high-quality paper jogger that can process as many as 1,000 sheets at once? If it does, you'll want to check out the MBM AJ 700. It's an air-feed jogger that has a lot of great features and can definitely withstand a lot of use. Here's an in-depth look at this device. The AJ 700 can handle various sizes of paper. The smallest pieces it can handle mature 5" x 7.375" (just slightly smaller than letter-size) while the largest measure 12.875" x 17.3/4" (slightly larger than tabloid size). Since this device can handle so many sizes of paper, it's very versatile. This jogger has a halfhorsepower motor which makes it pretty powerful. The machine can process up to 1,000 sheets at a time (thanks in part to a tray that's 4" deep). That's equal to two entire reams. Its jogging capacity makes it perfect for high-output environments such as mail rooms, print
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shops, and more. The AJ 700 is an air-feed jogger. It's capable of forcing air between your sheets of paper so that they won't stick together due to a build-up of static electricity. This also helps get rid of any powder. Thus, when you run your paper through other devices (binding machines, folders, etc.) they'll run through the unit smoothly and you won't have to deal with a bunch of misfeeds. On this machine's great features is that it's able to jog items that have just been printed such as press sheets. This helps the ink dry much faster. This device can be operated by foot pedal so your hands will remain free. You can run this machine in one of two modes: vibrate- and air-only. There's an automatic timer for your convenience. The AJ 700 is a floor model unit so it's not suitable for desktop use. Its dimensions are 17" (depth) x 22" (width) x 45.25" (height), so it's pretty large. Although this machine is on the heavy side (it weighs 84 pounds), it's fairly easy to move around because there are casters on the bottom of it. If durability is important to, you'll be happy to know that the AJ 700 features all-metal construction. It's really one of the most durable joggers on the market. Finally, this jogger comes with a one-year warranty. Unfortunately, the coverage doesn't extends to the machine's wearables. As you can now see, the MBM AJ 700 is a terrific air-feed paper jogger. This machine has an excellent jogging capacity there are too many of these devices that can process up to 1,000 sheets at once. It's great that this machine can also handle varying sizes of paper. You'll be able to jog just about anything, even recently printed documents. One thing to keep in mind is that this is a pretty big device and that it sits on the floor, so it might not be suitable for all work environments. Also, it would nice if the warranty covered wearables. Nonetheless, the AJ 700 is an excellent jogger, so consider adding it to your workplace today. About the author: If you'd like to purchase the MBM AJ 700, you should really visit MyBinding.com. They have this product available at a great price and they also have a wide selection of other Paper Handling Equipment. Plus, you'll get free shipping on all orders over .00. If you'd like to learn more about these types of products, you can read all about them at MyBindingBlog.com. Check it out for yourself now! Source: http://www.articlesbase.com/small-business-articles/a-in-depth-look-at-the-mbm-aj-700-airfeed-paper-jogger-3253054.html
Infant Car Seat
Posted in Baby Jogger |
Tags: jogger
26 Responses to â&#x20AC;&#x153;MBM AJ 700 Air-Feed Paper Joggerâ&#x20AC;? N A: February 6, 2012 at 9:50 am What is the velocity of the jogger relative to the water? A large cruise boat is moving at 15km/h East relative to the water. A person jogging on the ship moves across the ship in a northerly direction at 6km/h.What is the velocity of the jogger relative to the water? Reply
Elijah W: February 12, 2012 at 5:28 am
February 12, 2012 at 5:28 am just add the two velocities together because newtonian physics works at relatively low speeds. Reply
onekawaiikitty: February 13, 2012 at 9:03 am What is the magnitude of the joggerâ&#x20AC;&#x2122;s displacement, start to finish? A city jogger runs four blocks due east, eight blocks due south, and another two blocks due east. Assume all blocks are equal of size. What is the magnitude of the joggerâ&#x20AC;&#x2122;s displacement, start to finish? Reply
OldPilot: February 24, 2012 at 9:23 am 4 + 2 = 6 blocks East 8 blocks South Pythagorean Theorem 6^2 + 8^2 = D^2 36 + 64 = D^2 D^2 = 100 D = 10 units Reply
Anonymous: February 13, 2012 at 9:30 pm A jogger travels a route that has two parts. The first is a displacement A of 3.14 km due south, and the secon? A jogger travels a route that has two parts. The first is a displacement A of 3.14 km due south, and the second involves a displacement B that points due east. If the resultant displacement A+B has magnitude 3.77 km, what is the magnitude of B? Reply
Vivek S: February 19, 2012 at 12:18 pm if u look at ur question nd make diagram of path it will turn out to be a RIGHT ANGLE TRIANGLE in this the hypotenuse along with one side is given . u have to find the other side . Now i believe that i made ur problem clear to u nd the solution will be in understandable form New hot app:
we know –> a^2 + b^2 = c^2 ( where a nd b r sides while c is hypotenuse of right angled triangle ). (3.14)^2 + b^2 = (3.77)^2 b = 14.2129 – 9.8596 b^2 = 4.3533 b = 2.0864 ( ur answer) therefore magnitude of B is 2.0864 . i hope u understand the method. thnx Reply
JaneK: February 16, 2012 at 12:05 pm what do I need to pay attention to a swivel jogger before buying? The swivel jogger is more and more popular in the world. I just want to know what I need to pay attention to a swivel jogger before buying. I really want to get one but I don’t know how to choice. Please give me some suggestions if you have the experiences. Reply
Jacky L: February 17, 2012 at 4:50 am Check the instructions of which ever models you are looking at. Some of these three wheel models are not meant for jogging. For instance, the Jeep three wheeler plainly says not to use it as a jogger. The swivel wheel is not stable enough for jogging. Reply
Anonymous: February 16, 2012 at 7:04 pm A jogger jogs around a circular track with a diameter of 325 m in 13 minutes. What was the jogger’s average sp? A jogger jogs around a circular track with a diameter of 325 m in 13 minutes. What was the jogger’s average speed in m/s? Reply
Anonymous: February 28, 2012 at 11:11 am You need to convert the Diameter to the circumference. Which is 510.5088….. 0.51088km in 13m =39 m/s. Basic maths. Remember look at the key words of the problem. Maths can be fun
the problem. Maths can be fun Reply
m6: February 17, 2012 at 3:43 pm How will the velocity components found in part (a) change if the jogger’s speed is halved? A jogger runs with a speed of 3.50m/s in a direction 35.0 degree above the X axis. A.) Find the x and y components of the jogger’s velocity. B.)How will the velocity components found in part (a) change if the jogger’s speed is halved? Reply
J9900: March 4, 2012 at 3:45 pm Draw an angle that’s 35 degrees above the x-axis and label it 3.5m/s. Draw a vertical line and a horizontal line from the starting point of the angle. Label the horizontal line “x.” Draw another vertical line to the right so that a triangle is formed. Label the new vertical line “y.” Using trig, sin = opposite/hypotenuse, so sin35 = y/3.5, and solving for y gives y = 3.5sin35. Using trig, cos = adjacent/hypotenuse, so cos35 = x/3.5, and solving for x gives x = 3.5cos35. If the jogger’s speed is halved, then 3.5 becomes 1.75, so y = 1.75sin35, and x = 1.75cos35. Reply
Anonymous: February 17, 2012 at 10:36 pm How much water evaporates from jogger’s skin each hour when heat energy is at a rate of 878.2W? A jogger generates heat energy at a rate of 878.2 W. If all this energy is removed by sweating, how much water (in kg) must evaporate from the jogger’s skin each hour? Reply
Paul: February 27, 2012 at 10:32 am Water’s latent heat of evaporation is 2,270 kJ/kg. A watt is a J/s . That’s all you need. Reply
Anonymous:
Anonymous: February 25, 2012 at 7:55 pm If a jogger is moving at a constant speed are they accelerating? Homework question: Which of the following objects is not accelerating? A. a jogger moving at a constant speed B. a car slowing down C.earth orbiting the sun D. a car speeding up Reply
Raymond: February 28, 2012 at 10:34 pm In the “frame of reference” of this homework question, the answer is A. A better wording would have been to have the jogger go at a constant speed without changing direction (constant “velocity”). Otherwise, the Earth can be considered as orbiting the Sun at a constant speed (at least, as constant as the speed of a jogger). But everyone knows that the Earth’s orbit “turns” which implies a sideways acceleration. If you really, really wanted to be nit-picking, you could say that the jogger in running on Earth’s surface, which is curved, so that the jogger’s direction changes by 1/40,000 of a circle (0.009 degrees) for each kilometre he runs in a supposed straight line. The downward acceleration he needs for this is provided by gravity (just like Earth’s curve around the Sun). But, unless you feel like getting into a very complicated argument with the teacher, just answer A and take the points. — B the car is decelerating, which is simply an acceleration with a negative sign (mathematically, it is still an acceleration). C we have discussed above (sideways acceleration) D “speeding up” is just a crude way of saying “accelerating” (crude in the sense that using the phrase “speeding up” implies that you do not have a clear value of acceleration, you just know it is accelerating). Reply
mrchangmitsubishi: February 27, 2012 at 6:27 am A jogger jogs around a circular track with a diameter of 300 m in 10 minutes. What was the average speed m/s? A jogger jogs around a circular track with a diameter of 300 m in 10 minutes. What was the joggers average speed in m/s? Reply
╓─────╖ ║ Donny ║ ╙─────╜: March 3, 2012 at 9:48 pm 10 minutes = 10 * 60 = 600 seconds, which means the jogger runs at a speed of 300/600 m/s = 0.5 m/s
of 300/600 m/s = 0.5 m/s Hope that helps! Reply
Anonymous: February 27, 2012 at 9:57 pm What angular distance does the jogger cover in each of the following units? A jogger on a circular track that has a radius of 0.280 km runs a distance of 2.00 km. What angular distance does the jogger cover in each of the following units? (a) radians (b) degrees Reply
DH: February 28, 2012 at 4:45 pm It is unclear if you mean in one revolution. Assuming that then a) each revolution is 2*pi rad = 6.28 rad b) 360o If this is not what’s meant by the problem please re-post Reply
Anonymous: March 4, 2012 at 5:13 pm How would you estimate the speed of a jogger and a snowflake? This is for summer homework and ive never taken physics before so i don’t know the answer, and neither do my friends. Thank you in advance. Oh, and its a jogger and a snowflake separately. I combined two questions in one. Reply
Anonymous: March 6, 2012 at 5:06 pm Altho the prior answer is correct in the fundamental meaning of speed (ie distance/time). It may be that the question focus is on the word “Estimation” – which is different from taking actual measurements as with a stopwatch and tape measure. Estimation is to use your experience and what we call an “educated guess”. So for the runner you *estimate* how far he travels and how much time it took. You can “count time” by the “1000 and…” method, and use surroundings of known objects like: fence posts, sidewalk expansion cracks, and telephone poles (are they 100 ft apart?), etc. Same for the snowflake altho you’ll have move to a somewhat colder climate in summer to actually perform this estimation at this time of year :>) Reply
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Anonymous: March 4, 2012 at 9:35 pm If all this energy is removed by sweating, how much water must evaporate from the jogger’s skin each hour? A jogger generates heat energy at a rate of 815.3 W. If all this energy is removed by sweating, how much water must evaporate from the jogger’s skin each hour? Reply
DH: March 9, 2012 at 9:30 pm Water evaporates from the skin at less than 100oC, so the amount of mass required comes from Q = m*Lv where Lv is the latent heat of vaporization which for sweat at 37oC = 2.426×10^6J/kg Now Q = P*t = 815.3W*3600s = 2935080J So m = Q/Lv = 2935080/2.426×10^6 = 1.21kg Reply
Danielle: March 6, 2012 at 3:43 pm What is a good name for my World Geography mind jogger group? I am in 9th grade and it is a group of 4 fairly smart students and mind jogger is a way we review for the test…any catchy names? thank you so much if you can help? Reply
picador: March 7, 2012 at 2:07 pm How about “The Navigators”? Reply
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