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Unformatted text preview: Agha (sa9896) HW6 bohm (59970) 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A 13994 N car traveling at 55.1 km/h rounds a curve of radius 1 . 39 10 2 m. The acceleration of gravity is 9 . 81 m / s 2 . a) Find the centripetal acceleration of the car. Correct answer: 1 . 68532 m / s 2 . Explanation: Basic Concept: a c = v t 2 r Given: r = 1 . 39 10 2 m v t = 55 . 1 km / h Solution: a c = (55 . 1 km / h) 2 1 . 39 10 2 m parenleftbigg 1000 m 1 km parenrightbigg 2 parenleftbigg 1 h 3600 s parenrightbigg 2 = 1 . 68532 m / s 2 002 (part 2 of 3) 10.0 points b) Find the force that maintains circular mo tion. Correct answer: 2404 . 12 N. Explanation: Basic Concepts: F c = ma c F g = mg Given: F g = 13994 N g = 9 . 81 m / s 2 Solution: F c = parenleftbigg F g g parenrightbigg a c = parenleftbigg 13994 N 9 . 81 m / s 2 parenrightbigg (1 . 68532 m / s 2 ) = 2404 . 12 N 003 (part 3 of 3) 10.0 points c) Find the minimum coefficient of static fric tion between the tires and the road that will allow the car to round the curve safely. Correct answer: 0 . 171797. Explanation: Basic Concept: F c = F s = s F n = s F g Solution: s = F c F g = 2404 . 12 N 13994 N = 0 . 171797 004 10.0 points A(n) 0 . 59 kg object is swung in a vertical...
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This note was uploaded on 11/03/2008 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Kaplunovsky

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