h17 - hernandez (ejh742) homework 17 Turner (58120) This...

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hernandez (ejh742) – homework 17 – Turner – (58120) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A 1640 kg car starts From rest and accelerates uniFormly to 15 . 9 m / s in 15 . 5 s . Assume that air resistance remains con- stant at 337 N during this time. ±ind the average power developed by the engine. Correct answer: 21 . 5196 hp. Explanation: m = 1640 kg , v i = 0 m / s , v f = 15 . 9 m / s , and Δ t = 15 . 5 s . The acceleration oF the car is a = v f - v i Δ t = v f Δ t since v i = 0, so a = 15 . 9 m / s 15 . 5 s = 1 . 02581 m / s 2 . Thus the constant Forward Force due to the engine is Found From s F = F engine - F air = ma F engine = F air + ma = 337 N + (1640 kg) ( 1 . 02581 m / s 2 ) = 2019 . 32 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 p v f 2 P = (2019 . 32 N) ± 15 . 9 m / s 2 ²± 1 hp 764 W ² = 21 . 5196 hp . 002 (part 2 oF 2) 10.0 points ±ind the instantaneous power output oF the engine at t = 15 . 5 s just beFore the car stops accelerating. Correct answer: 43 . 0392 hp. Explanation: The instantaneous velocity is 15 . 9 m / s and the instantaneous power output oF the engine is P = F engine v f = (2019 . 32 N)(15 . 9 m / s) ± 1 hp 764 W ² = 43 . 0392 hp . 003 (part 1 oF 2) 10.0 points Three 7 kg masses are located at points in the xy plane as shown. 34 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: 3 . 0425 × 10 8 N. Explanation: Let : m = 7 kg , x = 34 cm = 0 . 34 m , y = 54 cm = 0 . 54 m , and G = 6 . 6726 × 10 11 N · m 2 / kg 2 .
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hernandez (ejh742) – homework 17 – Turner – (58120) 2 The force from the mass on the right points
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h17 - hernandez (ejh742) homework 17 Turner (58120) This...

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