solution18

# solution18 - fullwood (tf5349) Homework 7 Uniform Circular...

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fullwood (tf5349) – Homework 7 Uniform Circular MOtion adn Gravitation – Catala – (49033) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. 001 (part 1 of 3) 10.0 points An air puck of mass 0 . 365 kg is tied to a string and allowed to revolve in a circle of radius 0 . 58 m on a horizontal, frictionless table. The other end of the string passes through a hole in the center of the table and a mass of 1 . 3 kg is tied to it. The suspended mass remains in equilibrium while the puck revolves. The acceleration of gravity is 9 . 8 m / s 2 . 0 . 58 m v 1 . 3 kg 0 . 365 kg What is the tension in the string? Correct answer: 12 . 74 N. Explanation: r v Mg m Since the suspended mass is in equilibrium, the tension is T = M g = (1 . 3 kg) ( 9 . 8 m / s 2 ) = 12 . 74 N . 002 (part 2 of 3) 10.0 points What is the horizontal force acting on the puck? Correct answer: 12 . 74 N. Explanation: The horizontal force acting on the puck is the tension in the string, so F c = T = 12 . 74 N . 003 (part 3 of 3) 10.0 points What is the speed of the puck? Correct answer: 4 . 49938 m / s. Explanation: F c = mv 2 r v = r F c r m = R (12 . 74 N) (0 . 58 m) 0 . 365 kg = 4 . 49938 m / s . 004 (part 1 of 2) 10.0 points Because of Earth’s rotation about its axis, a point on the Equator experiences a cen- tripetal acceleration of 0 . 034 m / s 2 , while a point at the poles experiences no centripetal acceleration. What is the apparent weight at the equator of a person having a mass of 130 . 5 kg? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1274 . 46 N. Explanation: Let : a c = 0 . 034 m / s 2 and m = 130 . 5 kg . Since the centripetal acceleration of a per- son is inward (toward the Earth’s axis), F c = mg - W a W a = mg - ma c = m ( g - a c ) = (130 . 5 kg) ( 9 . 8 m / s 2 - 0 . 034 m / s 2 ) = 1274 . 46 N .

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fullwood (tf5349) – Homework 7 Uniform Circular MOtion adn Gravitation – Catala – (49033) 2 005 (part 2 of 2) 10.0 points What is his apparent weight at the poles? Correct answer: 1278 . 9 N. Explanation: At the poles, the rotating velocity is zero, so F c = 0 = W a - mg W a = mg = (130 . 5 kg) ( 9 . 8 m / s 2 ) = 1278 . 9 N . 006 (part 1 of 5) 10.0 points A car travels at a speed of 30 m / s around a curve of radius 110 m. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 3 Mg μ 23 What is the net centripetal force needed to keep the car from skidding sideways? Correct answer: 10636
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## This note was uploaded on 06/10/2010 for the course PHY 100 taught by Professor Pro.kiroshima during the Spring '10 term at Adelphi.

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solution18 - fullwood (tf5349) Homework 7 Uniform Circular...

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