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Unformatted text preview: Le (tl8426) – 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 15 . 7 m at 1 . 46 s and at 3 . 33 m at 3 . 57 s. Find its displacement during this time in terval. Correct answer: − 12 . 37 m. Explanation: Let : x f = 3 . 33 m and x i = 15 . 7 m . The displacement is Δ x = x f − x i = 3 . 33 m − 15 . 7 m = − 12 . 37 m . 002 (part 2 of 3) 10.0 points What is its average velocity during this time interval? Correct answer: − 5 . 86256 m / s. Explanation: Let : t i = 1 . 46 s and t f = 3 . 57 s . The average velocity is ¯ v = Δ x Δ t = x f − x i t f − t i = 3 . 33 m − 15 . 7 m 3 . 57 s − 1 . 46 s = − 5 . 86256 m / s . 003 (part 3 of 3) 10.0 points Calculate the particle’s average speed during this time interval. Correct answer: 5 . 86256 m / s. Explanation: The average speed is the magnitude of the average velocity: speed avg =  ¯ v  = − 5 . 86256 m / s  = 5 . 86256 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 . 3 s for the pot to fall past his window, which is 10 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: 51 . 7419 m. Explanation: Let : Δ t win = 0 . 3 s , Δ h win = 10 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 + Δ t win ) 2 , where Δ t win = t bottom − t top . Δ y win = 1 2 g [( t top + Δ t win ) 2 − t 2 top ] = 1 2 g (2 t top Δ t win ) + 1 2 g (Δ t win ) 2 t top = Δ y win g Δ t win − 1 2 Δ t win = 10 m (9 . 81 m / s 2 ) (0 . 3 s) − 1 2 (0 . 3 s) = 3 . 24789 s and Le (tl8426) – HW12 – gentle – (56245) 2 y top = 1 2 (9 . 81 m / s 2 ) (3 . 24789 s) 2 = 51 . 7419 m . 005 (part 1 of 2) 10.0 points The coefficient of static friction between a rubber tire and the road surface is 0 . 45. The acceleration of gravity is 9 . 81 m / s 2 . What is the maximum acceleration of a 880 kg fourwheeldrive truck if the road makes an angle of 18 ◦ with the horizontal and the truck is climbing? Correct answer: 1 . 16698 m / s 2 . Explanation: Let : μ s = 0 . 45 , m = 880 kg , and θ = 18 ◦ ....
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This note was uploaded on 12/05/2011 for the course PHY 301 taught by Professor Swinney during the Fall '07 term at University of Texas.
 Fall '07
 Swinney
 mechanics

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