# W l 2 cos θ 015 100points a student weighing 700 n

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W L 2 cos θ 015 10.0points A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the student to overcome gravity is most nearly 1. P = 87 . 5 W. 2. P = 1 . 1 W. 3. P = 5 , 600 W. 4. P = 560 W. correct 5. P = 875 W. Explanation: P = W t = F d t = (700 N)(8 m) 10 s = 560 W . 016 10.0points A 15 kg piece of metal (specific heat of 0 . 05 cal / g · C) having a temperature of 65 C is added to 0 . 8 kg of water having a tempera- ture of 7 C. What is the final equilibrium temperature of the system? Assume there is no loss of heat to the surroundings. The specific heat of water is 1 cal / g · C. 1. 46.8235 2. 64.9333 3. 37.4848 4. 35.0645 5. 39.4545 6. 33.6667 7. 54.5 8. 41.4 9. 43.8913 10. 23.0 Correct answer: 35 . 0645 C. 1. There is no change in the internal energy of the gas. correct 2. There is no volume change undergone by the gas. 3. None of these is true. 4. There is no work done by the gas. Explanation: Let : m 1 = 15 kg , c 1 = 0 . 05 cal / g · C , T 1 = 65 C , m 2 = 0 . 8 kg , c 2 = 1 cal / g · C , and T 2 = 7 C . There is no loss of heat to the surroundings, so m 1 c 1 ( T 1 - T f ) = m 2 c 2 ( T f - T 2 ) m 1 c 1 T 1 - m 1 c 1 T f = m 2 c 2 T f - m 2 c 2 T 2 m 1 c 1 T 1 + m 2 c 2 T 2 = ( m 2 c 2 + m 1 c 1 ) T f T f = m 1 c 1 T 1 + m 2 c 2 T 2 m 1 c 1 + m 2 c 2 Since m 1 c 1 T 1 + m 2 c 2 T 2 = (15 kg) (0 . 05 cal / g · C) (65 C) + (0 . 8 kg) (1 cal / g · C) (7 C) = 54 . 35 kg · cal / g and m 1 c 1 + m 2 c 2 = (15 kg) (0 . 05 cal / g · C) + (0 . 8 kg) (1 cal / g · C) = 1 . 55 kg · cal / g · C , then the final temperature will be T f = 54 . 35 kg · cal / g 1 . 55 kg · cal / g · C = 35 . 0645 C . 017 10.0points In an isothermal process working on an ideal gas,
Version 025 – Final – yao – (92695) 8 5. There is no heat exchange during the process. Explanation: The temperature is constant in an isother- mal process. The internal energy of an ideal gas consists only of the kinetic energy of its molecules, which, in turn, depends only on the temperature. Thus, there is no change in the internal energy of an ideal gas. 018 10.0points A circular disk with moment of inertia 1 2 m R 2 , mass m and radius R is mounted at its center, about which it can rotate freely. A light cord wrapped around it supports a weight m g . R I ω m T g Find the total kinetic energy of the system when the weight is moving at a speed v . 8. K = 3 4 m v 2 correct 9. K = 5 2 m v 2 Explanation: v = r ω , so the total kinetic energy is K tot = K m + K rot = 1 2 m v 2 + 1 2 I ω 2 = 1 2 m v 2 + 1 2 parenleftbigg 1 2 m R 2 parenrightbigg parenleftBig v R parenrightBig 2 = 3 4 m v 2 . 019 10.0points