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Unformatted text preview: Answer, Key Homework 4 David McIntyre 45123 Mar 25, 2004 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. Chapters 3 and 4 problems. 001 (part 1 of 1) 0 points A ball on the end of a string is whirled around in a horizontal circle of radius 0 . 454 m. The plane of the circle is 1 . 49 m above the ground. The string breaks and the ball lands 2 . 71 m away from the point on the ground directly beneath the balls location when the string breaks. The acceleration of gravity is 9 . 8 m / s 2 . Find the centripetal acceleration of the ball during its circular motion. Correct answer: 53 . 1977 m / s 2 . Explanation: In order to find the centripetal acceleration of the ball, we need to find the initial velocity of the ball. Let y be the distance above the ground. After the string breaks, the ball has no initial velocity in the vertical direction, so the time spent in the air may be deduced from the kinematic equation, y = 1 2 gt 2 Solving for t , t = r 2 y g Let d be the distance traveled by the ball. Then v x = d t = d q 2 y g Hence, the centripetal acceleration of the ball during its circular motion is a c = v 2 x r = d 2 g 2 yr = 53 . 1977 m / s 2 002 (part 1 of 2) 0 points The orbit of a Moon about its planet is ap- proximately circular, with a mean radius of 2 . 44 10 8 m. It takes 26 . 4 days for the Moon to complete one revolution about the planet. Find the mean orbital speed of the Moon. Correct answer: 672 . 128 m / s. Explanation: Dividing the length C = 2 r of the trajectory of the Moon by the time T = 26 . 4 days = 2 . 28096 10 6 s of one revolution (in seconds!), we obtain that the mean orbital speed of the Moon is v = C T = 2 r T = 2 (2 . 44 10 8 m) 2 . 28096 10 6 s = 672 . 128 m / s . 003 (part 2 of 2) 0 points Find the Moons centripetal acceleration. Correct answer: 0 . 00185146 m / s 2 . Explanation: Since the magnitude of the velocity is con- stant, the tangential acceleration of the Moon is zero. For the centripetal acceleration we use the formula a c = v 2 r = (672 . 128 m / s) 2 2 . 44 10 8 m = 0 . 00185146 m / s 2 . 004 (part 1 of 3) 0 points A train slows down at a constant rate as it rounds a sharp circular horizontal turn. Its initial speed is not known. It takes 16 . 4 s to slow down from 78 km / h to 30 km / h. The radius of the curve is 180 m....
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