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# cw9 - patel(kp9583 CW9 Mackie(20211 This print-out should...

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patel (kp9583) – CW9 – Mackie – (20211) 1 This print-out should have 3 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An Olympic skier moving at 30 m / s down a 19 slope encounters a region of wet snow of coefficient of kinetic friction μ = 0 . 59. The acceleration of gravity is 9 . 8 m / s 2 . How far down the slope does she travel before coming to a halt? Correct answer: 197 . 679 m. Explanation: Align the x axis along the incline, and let down the incline be positive. The normal force is N = m g cos θ . Hence the frictional force is f = μ N = μ m g cos θ Applying Newton’s second law to the skier along the incline yields summationdisplay F = m g sin θ - μ m g cos θ = m a , or solving for a , a = g [sin θ - μ cos θ ] = (9 . 8 m / s 2 ) × [sin 19 - (0 . 59) cos 19 ] = - 2 . 27642 m / s 2 . To find how far the slope the skier travels before stopping, use the kinematic equation v 2 = v 2 0 + 2 a d . Setting v = 0 and solving for d , d = - v 2 0 2 a = - (30 m / s) 2 2 ( - 2 . 27642 m / s 2 ) = 197 . 679 m . 002 (part 1 of 2) 10.0 points An air puck of mass 0.032 kg is tied to a string and allowed to revolve in a circle of radius 1.2

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