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Unformatted text preview: Pickering, Tracey Homework 15 Due: Feb 27 2006, noon Inst: Drummond 1 This print-out should have 12 questions. Multiple-choice 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 = g 2. P = mg h t correct 3. P = mg h 4. P = mg ht 5. P = mh 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 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 200 W light- bulb that is lit for 5 . 6 hours? Correct answer: 93723 . 8 h. Explanation: Let : P = 3 cal / min and t = 5 . 6 h . The energy expended by the light in 5 . 6 h is E = P t = (200 W) (5 . 6 h) 3600 s h = 4 . 032 10 6 J . The rate at which energy is being used while walking is P = (3 cal / min) (0 . 239 J / cal) = 0 . 717 J / min , so the time need to produce the same amount of energy is t = E P = 4 . 032 10 6 J . 717 J / min 1 h 60 min = 93723 . 8 h . 003 (part 1 of 2) 10 points Find the power input of a force ~ F = (9 N) + (6 N) k acting on a particle that moves with a velocity ~v = (1 m / s) . Correct answer: 9 W. Explanation: Let : ~ F = (9 N) + (6 N) k and ~v = (1 m / s) ....
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This note was uploaded on 06/23/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
- Fall '08