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# test2 - Version 072 TEST02 tsoi(92940 This print-out should...

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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 10.0points 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 10.0points 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 car’s 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 .

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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 = μ s N = μ s ( M g ) . Consequently, the net horizontal acceleration of the car is limited by M | a | ≤ μ s M g | a | ≤ μ s g . When the car starts to exceed this accelera- tion limit, the tires skid, so the static friction coefficient is μ s = | a | max g = 5 . 33471 m / s 2 9 . 8 m / s 2 = 0 . 544358 . 003 10.0points The system of point masses shown in the figure is rotating at an angular speed of ω = 1 . 41 rev / s. The masses (all equal) are connected by light, flexible spokes that can be lengthened or shortened. At the beginning, each spoke is r = 0 . 879 m long.
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