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HW17_Solutions - hiarker(srh959 homework 17 Turner(58220...

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hiarker (srh959) – homework 17 – Turner – (58220) 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 (part 1 of 2) 10.0 points A 1100 kg car starts from rest and accelerates uniformly to 12 . 7 m / s in 14 . 6 s . Assume that air resistance remains con- stant at 361 N during this time. Find the average power developed by the engine. Correct answer: 11 . 2176 hp. Explanation: m = 1100 kg , v i = 0 m / s , v f = 12 . 7 m / s , and Δ t = 14 . 6 s . The acceleration of the car is a = v f - v i Δ t = v f Δ t since v i = 0, so a = 12 . 7 m / s 14 . 6 s = 0 . 869863 m / s 2 . Thus the constant forward force due to the engine is found from summationdisplay F = F engine - F air = m a F engine = F air + m a = 361 N + (1100 kg) ( 0 . 869863 m / s 2 ) = 1317 . 85 N . The average velocity of the car during this interval is v av = v f + v i 2 , so the average power output is P = F engine v av = F engine parenleftBig v f 2 parenrightBig = (1317 . 85 N) parenleftbigg 12 . 7 m / s 2 parenrightbigg parenleftbigg 1 hp 764 W parenrightbigg = 11 . 2176 hp . 002 (part 2 of 2) 10.0 points Find the instantaneous power output of the engine at t = 14 . 6 s just before the car stops accelerating. Correct answer: 22 . 4352 hp. Explanation: The instantaneous velocity is 12 . 7 m / s and the instantaneous power output of the engine is P = F engine v f = (1317 . 85 N)(12 . 7 m / s) parenleftbigg 1 hp 764 W parenrightbigg = 22 . 4352 hp . 003 (part 1 of 2) 10.0 points Three 2 kg masses are located at points in the xy plane as shown. 43 cm 54 cm What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? The universal gravita- tional constant is 6 . 6726 × 10 11 N · m 2 / kg 2 . Correct answer: 1 . 70924 × 10 9 N. Explanation: Let : m = 2 kg , x = 43 cm = 0 . 43 m , y = 54 cm = 0 . 54 m , and G = 6 . 6726 × 10 11 N · m 2 / kg 2 .
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hiarker (srh959) – homework 17 – Turner – (58220) 2 The force from the mass on the right points in the x direction and has magnitude F 1 = G m m x 2 = G m 2 x 2 = (6 . 6726 × 10 11 N · m 2 / kg 2 ) (2 kg) 2 (0 . 43 m) 2 = 1 . 4435 × 10 9 N .
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