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Unformatted text preview: lee (cyl327) HW12 gentle (56245) 1 This printout should have 50 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. This is a set of review prolems. 001 (part 1 of 3) 10.0 points A particle moving uniformly along the x axis is located at 9 . 18 m at 1 . 6 s and at 3 . 19 m at 3 . 54 s. Find its displacement during this time in terval. Correct answer: 5 . 99 m. Explanation: Let : x f = 3 . 19 m and x i = 9 . 18 m . The displacement is x = x f x i = 3 . 19 m 9 . 18 m = 5 . 99 m . 002 (part 2 of 3) 10.0 points What is its average velocity during this time interval? Correct answer: 3 . 08763 m / s. Explanation: Let : t i = 1 . 6 s and t f = 3 . 54 s . The average velocity is v = x t = x f x i t f t i = 3 . 19 m 9 . 18 m 3 . 54 s 1 . 6 s = 3 . 08763 m / s . 003 (part 3 of 3) 10.0 points Calculate the particles average speed during this time interval. Correct answer: 3 . 08763 m / s. Explanation: The average speed is the magnitude of the average velocity: speed avg =  v  =  3 . 08763 m / s  = 3 . 08763 m / s . 004 10.0 points A flowerpot falls from the ledge of an apart ment building. A person in an apartment below, coincidentally holding a stopwatch, no tices that it takes 0 . 6 s for the pot to fall past his window, which is 14 m high. How far above the top of the window is the ledge from which the pot fell? The accelera tion of gravity is 9 . 81 m / s 2 . Correct answer: 21 . 1909 m. Explanation: Let : t win = 0 . 6 s , h win = 14 m , and g = 9 . 81 m / s 2 . Using a constantacceleration equation, ex press the distance y below the ledge from which the pot fell as a function of time: y = y + v + 1 2 g t 2 = 1 2 g t 2 since v = y = 0. The distance to the top is y top = 1 2 g t 2 top and to the bottom y bottom = 1 2 g ( t top + t win ) 2 , where t win = t bottom t top . y win = 1 2 g [( t top + t win ) 2 t 2 top ] = 1 2 g (2 t top t win ) + 1 2 g ( t win ) 2 t top = y win g t win 1 2 t win = 14 m (9 . 81 m / s 2 ) (0 . 6 s) 1 2 (0 . 6 s) = 2 . 07853 s and lee (cyl327) HW12 gentle (56245) 2 y top = 1 2 (9 . 81 m / s 2 ) (2 . 07853 s) 2 = 21 . 1909 m . 005 (part 1 of 2) 10.0 points The coefficient of static friction between a rubber tire and the road surface is 0 . 55. The acceleration of gravity is 9 . 81 m / s 2 . What is the maximum acceleration of a 950 kg fourwheeldrive truck if the road makes an angle of 14 with the horizontal and the truck is climbing? Correct answer: 2 . 86198 m / s 2 . Explanation: Let : s = 0 . 55 , m = 950 kg , and = 14 ....
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This note was uploaded on 02/28/2012 for the course PHY 316 taught by Professor Markert during the Fall '10 term at University of Texas at Austin.
 Fall '10
 MARKERT
 mechanics

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