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# hw17 - hinojosa(jlh3938 homework 17 Turner(58185 This...

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hinojosa (jlh3938) – homework 17 – Turner – (58185) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A hot rod of mass 1100 kg, starting from rest reaches a speed of 150 m / s in only 25 . 4 s. What is the average output power? 1. 0 . 974409 MW 2. 0 . 000127875 MW 3. 0 . 00649606 MW 4. 0 . 0191813 MW 5. 0 . 487205 MW correct Explanation: The average output power is equal to the kinetic energy acquired divided by the total time, so P = 1 2 mv 2 t = (1100 kg) (150 m / s) 2 2 (25 . 4 s) · mW 10 6 W = 0 . 487205 MW 002 10.0 points A car with a mass of 1 . 74 × 10 3 kg starts from rest and accelerates to a speed of 17.3 m/s in 12.0 s. Assume that the force of resistance remains constant at 308.1 N during this time. What is the average power developed by the car’s engine? Correct answer: 24363 . 6 W. Explanation: Let : m = 1 . 74 × 10 3 kg , v f = 17 . 3 m / s , Δ t = 12 . 0 s , and F r = 308 . 1 N . v f = a Δ t and Δ x = 1 2 a t ) 2 since v i = 0 m/s, and F = m a + F r , so P = F Δ x Δ t = F · 1 2 a t ) 2 Δ t = F a Δ t 2 = ( m a + F r ) v f 2 = parenleftBig m v f Δ t + F r parenrightBig · v f 2 = bracketleftbigg (1740 kg)(17 . 3 m / s) 12 s + 308 . 1 N bracketrightbigg × 17 . 3 m / s 2 = 24363 . 6 W . 003 10.0 points A car weighing 7400 N moves along a level highway with a speed of 100 km / h. The power of the engine at this speed is 71 kW. The car encounters a hill inclined at an angle of 6 . 5 with respect to the horizontal. If there is no change in the power of the engine, and no change in the resistive forces acting on the car, what is the new speed of the car on the hill? Correct answer: 75 . 3159 km / h. Explanation: Since the velocity is constant, the car must be in equilibrium. Assume F is the pulling force provided by the engine and f is the fric- tional force. On the level highway F 1 = f F 1 P v On the hill: F 2 = f + mg sin( θ ) v 2 P F 2 = F 1 v F 2 = F 1

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