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Unformatted text preview: min (mm64262) – Homework 6 – lee – (GEDB00864) 1 This print-out should have 7 questions. Multiple-choice 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 64 cm apart on the same axle. From the angular displacement 25 . 6 ◦ of the two bul- let holes in the disks and the rotational speed 1403 rev / min of the disks, we can determine the speed of the bullet. 25 . 6 ◦ v 1403 rev / min 64 cm What is the speed of the bullet? Correct answer: 210 . 45 m / s. Explanation: Let : ω = 1403 rev / min , d = 64 cm , and θ = 25 . 6 ◦ . θ = ω t t = θ ω , so the speed of the bullet is v = d t = d ω θ = (64 cm) (1403 rev / min) 25 . 6 ◦ × 360 ◦ 1 rev 1 m 100 cm 1 min 60 s = 210 . 45 m / s . keywords: 002 10.0 points A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel of radius 0 . 4 m and observes that drops of water fly off tangentially. She mea- sures the height reached by drops moving ver- tically. A drop that breaks loose from the tire on one turn rises 38 . 7 cm above the tangent point. A drop that breaks loose on the next turn rises 35 cm above the tangent point (the angular speed of the wheel is decreasing). h . 4 m radius ω Find the angular deceleration of the wheel. The acceleration of gravity is 9 . 8 m / s 2 . As- sume the angular deceleration is constant. Correct answer: 0 . 360685 rad / s 2 . Explanation: Let : r = 0 . 4 m , h 1 = 38 . 7 cm = 0 . 387 m , h 2 = 35 cm = 0 . 387 m , and Δ θ = 1 rev = 2 π ....
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- Spring '11