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HW #7-solutions

# HW #7-solutions - wolz(cmw2833 HW#7 Antoniewicz(57420 This...

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wolz (cmw2833) – HW #7 – Antoniewicz – (57420) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points A 1560 kg car starts from rest and accelerates uniformly to 19 . 3 m / s in 18 . 1 s . Find the average power developed by the engine. Assume that air resistance remains constant at 382 N during this time. Correct answer: 26 . 4589 hp. Explanation: Let : m = 1560 kg , v i = 0 m / s , v f = 19 . 3 m / s , and Δ t = 18 . 1 s . The acceleration of the car is a = v f - v i Δ t = v f Δ t = 19 . 3 m / s 18 . 1 s = 1 . 0663 m / s 2 and 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 = 382 N + (1560 kg) ( 1 . 0663 m / s 2 ) = 2045 . 43 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 = (2045 . 43 N) parenleftbigg 19 . 3 m / s 2 parenrightbigg parenleftbigg 1 hp 746 W parenrightbigg = 26 . 4589 hp . 002(part2of2)10.0points Find the instantaneous power output of the engine at t = 18 . 1 s just before the car stops accelerating. Correct answer: 52 . 9178 hp. Explanation: The instantaneous velocity is 19 . 3 m / s and the instantaneous power output of the engine is P = F engine v f = (2045 . 43 N)(19 . 3 m / s) parenleftbigg 1 hp 746 W parenrightbigg = 52 . 9178 hp . 003(part1of2)10.0points On the way to the moon, the Apollo astro- nauts reach a point where the Moon’s gravi- tational pull is stronger than that of Earth’s. Find the distance of this point from the center of the Earth. The masses of the Earth and the Moon are 5 . 98 × 10 24 kg and 7 . 36 × 10 22 kg, respectively, and the distance from the Earth to the Moon is 3 . 84 × 10 8 m. Correct answer: 3 . 45653 × 10 8 m. Explanation: Let : M e = 5 . 98 × 10 24 kg , m m = 7 . 36 × 10 22 kg , and R e = 3 . 84 × 10 8 m . Consider the point where the two forces are equal. If r e is the distance from this point to the center of the Earth and r m is the distance from this point to the center of the Moon, then G m M e r 2 e = G m M m r 2 m r 2 m r 2 e = M m M e r m r e = radicalbigg M m M e .

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wolz (cmw2833) – HW #7 – Antoniewicz – (57420) 2 It is also true that R = r e + r m = r e + radicalbigg M m M e r e r e = R 1 + radicalbigg M m M e = 3 . 84 × 10 8 m 1 + radicalBigg 7 . 36 × 10 22 kg 5 . 98 × 10 24 kg = 3 . 45653 × 10 8 m . 004(part2of2)10.0points What is the acceleration due to the Earth’s gravity at this point? The value of the universal gravitational constant is 6 . 672 × 10 11 N · m 2 / kg 2 . Correct answer: 0 . 00333946 m / s 2 . Explanation: a = F m = G M e r 2 e a = (6 . 672 × 10 11 N · m 2 / kg 2 ) × 5 . 98 × 10 24 kg (3 . 45653 × 10 8 m) 2 = 0 . 00333946 m / s 2 . 005(part1of2)10.0points Given: G = 6 . 67259 × 10 11 N m 2 / kg 2 . A 1050 kg geosynchronous satellite or- bits a planet similar to Earth at a radius 1 . 96 × 10 5 km from the planet’s center. Its angular speed at this radius is the same as the rotational speed of the Earth, and so they ap- pear stationary in the sky. That is, the period of the satellite is 24 h .
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HW #7-solutions - wolz(cmw2833 HW#7 Antoniewicz(57420 This...

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