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Unformatted text preview: Version 072 TEST02 tsoi (92940) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Normally the rate at which you expend energy during a brisk walk is 3 . 1 Calories per minute. (A Calorie is the common unit of food energy, equal to 4184 . 1 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? 1. 1.7208 2. 2.18685 3. 0.834327 4. 1.99891 5. 7.35868 6. 1.29523 7. 3.54726 8. 6.6681 9. 1.69816 10. 1.27672 Correct answer: 1 . 27672 h. Explanation: Let : P = 3 . 1 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 = 9 . 936 10 5 J . The rate at which energy is being used while walking is P = (3 . 1 Cal / min) (4184 . 1 J / Cal) = 12970 . 7 J / min , so the time need to produce the same amount of energy is t = E P = 9 . 936 10 5 J 12970 . 7 J / min 1 h 60 min = 1 . 27672 h . 002 10.0 points To test the performance of its tires, a car travels along a perfectly flat (no banking) cir- cular track of radius 389 m. The car increases its speed at uniform rate of a t d | v | dt = 4 . 64 m / s 2 until the tires start to skid. If the tires start to skid when the car reaches a speed of 32 m / s, what is the coefficient of static friction between the tires and the road? The acceleration of gravity is 9 . 8 m / s 2 . 1. 0.355893 2. 0.63605 3. 0.582742 4. 0.323145 5. 0.544358 6. 0.559861 7. 0.509574 8. 0.620436 9. 0.406725 10. 0.759253 Correct answer: 0 . 544358. Explanation: Let : R = 389 m , v = 32 m / s , and g = 9 . 8 m / s 2 . We need the cars acceleration just before the skidding occurs. The car travels in a circle rather than along a straight line, so it has the centripetal acceleration a c = v 2 R . In addition, the car increases its speed, so there is a tangential acceleration a t d | v | dt = 4 . 64 m / s 2 . Version 072 TEST02 tsoi (92940) 2 The net acceleration of the car is the vector sum vector a = vector a c + vector a t ; since the tangential and the centripetal accel- erations have perpendicular orientations, the magnitude of the net acceleration is | a | = radicalBig a 2 c + a 2 t = radicalbigg v 4 R 2 + a 2 t = radicalBigg (32 m / s) 4 (389 m) 2 + (4 . 64 m / s 2 ) 2 = 5 . 33471 m / s 2 . The force providing this acceleration is the static friction force vector F s between the tires and the track the only horizontal force available on the flat track. The static friction force may have any horizontal direction, and any magnitude up to the maximum of F max s = s N = s ( M g ) ....
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test2 - Version 072 TEST02 tsoi (92940) 1 This print-out...

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