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Unformatted text preview: Morby, Grant – Homework 15 – Due: Feb 27 2006, noon – Inst: Drummond 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A weight lifter lifts a mass m at constant speed to a height h in time t . What is the average power output of the weight lifter? 1. ¯ P = mg ht 2. ¯ P = g 3. ¯ P = mh 4. ¯ P = mg h t correct 5. ¯ P = mg h Explanation: The amount of work to lift a mass m to a height h is mg h . The power is the amount of work done divided by the time t . Alternately, power has units of energy over time. The only answer with the correct units is ¯ P = mg h t . 002 (part 1 of 1) 10 points Normally the rate at which you expend energy during a brisk walk is 3 . 8 calories per minute. (A calorie is the common unit of food energy, equal to 0 . 239 J .) How long do you have to walk to produce the same amount of energy as a 40 W light bulb that is lit for 6 . 9 hours? Correct answer: 18233 . 9 h. Explanation: Let : P = 3 . 8 cal / min and t = 6 . 9 h . The energy expended by the light in 6 . 9 h is E = P t = (40 W) (6 . 9 h) × 3600 s h = 993600 J . The rate at which energy is being used while walking is P = (3 . 8 cal / min) (0 . 239 J / cal) = 0 . 9082 J / min , so the time need to produce the same amount of energy is t = E P = 993600 J . 9082 J / min · 1 h 60 min = 18233 . 9 h . 003 (part 1 of 2) 10 points Find the power input of a force ~ F = (4 N)ˆ ı + (2 N) ˆ k acting on a particle that moves with a velocity ~v = (5 m / s)ˆ ı . Correct answer: 20 W. Explanation: Let : ~ F = (4 N)ˆ ı + (2 N) ˆ k and ~v = (5 m / s)ˆ ı....
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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas at Austin.
 Spring '09
 KLEINMAN
 Physics, Work

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