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Unformatted text preview: valdez (vv689) Homework 13 florin (58140) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The speed of a moving bullet can be deter mined by allowing the bullet to pass through two rotating paper disks mounted a distance 75 cm apart on the same axle. From the angular displacement 23 . 4 of the two bul let holes in the disks and the rotational speed 1396 rev / min of the disks, we can determine the speed of the bullet. 23 . 4 v 1396 rev / min 75 cm What is the speed of the bullet? Correct answer: 268 . 462 m / s. Explanation: Let : = 1396 rev / min , d = 75 cm , and = 23 . 4 . = t t = , so the speed of the bullet is v = d t = d = (75 cm) (1396 rev / min) 23 . 4 360 1 rev 1 m 100 cm 1 min 60 s = 268 . 462 m / s . keywords: 002 10.0 points A bug is on the rim of a 78 rev / min, 12 in . diameter record. The record moves from rest to its final angular speed in 3 . 37 s. Find the bugs centripetal acceleration 1 . 5 s after the bug starts from rest. (1 in = 2.54 cm). Correct answer: 2 . 01444 m / s 2 . Explanation: Let : w = 78 rev / min , t = 3 . 37 s , r = 6 in , and t = 1 . 5 s . = t , so = t = t t = 78 rev / min 3 . 37 s (1 . 5 s) 1 min 60 s = 3 . 63567 rad / s , and a r = v 2 t r = r 2 = (6 in)(3 . 63567 rad / s) 2 1 cm 2 . 54 in 1 m 100 cm = 2 . 01444 m / s 2 . 003 10.0 points A small wheel of radius 1 . 6 cm drives a large wheel of radius 15 cm by having their circum ferences pressed together....
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 Spring '10
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