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# A clock pendulum oscillates at a frequency of 2.5 Hz. At t ISBN: 9780130606204 3

## Solution for problem 35P Chapter 11

Physics: Principles with Applications | 6th Edition

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Problem 35P

Problem 35P

A clock pendulum oscillates at a frequency of 2.5 Hz. At t = 0, it is released from rest starting at an angle of 15° to the vertical. Ignoring friction, what will be the position (angle) of the pendulum at (a) t = 0.25 s, (b) t = 1.60 s, and (c) t = 500 s? [Hint: Do not confuse the angle of swing θ of the pendulum with the angle that appears as the argument of the cosine.]

Step-by-Step Solution:

SolutionStep 1Explaining 1st and 2nd law of thermodynamics According to the 1st law of thermodynamics, the change in internal energy, U of a system is the resultant of heat added to the system, Q and work done by the system, W. That is, U = Q - W ------------ (1) The negative sign of work indicates that the work is done by the system. From the equation for 1st law, we can see that, this is obeying the principle of law of conservation of energy. According to the 2nd law of thermodynamics, the total entropy of an isolated system is increasing as the time increases. And from this law, we could explain the working principle of heat engines. So, the maximum possible efficiency of a heat engine will be the Carnot efficiency. So, the Carnot efficiency is, = Work done as output / Amount of heat extracted from hot reservoir That is, = W / Q --------------- (2) H But, we know that, the work done is, W = Q - Q -----H----C- (3) Where, Q - AmCnt of heat exhausted to cold reservoir So, rewriting this efficiency equation by using the new expression for work done as output. = (Q - Q ) / Q = 1 - (Q / Q ) ------------------ (4) H C H C H Or we can rewrite it in terms of temperature also. That is, = 1 - (T / C ) H----------------- (5)\n Normally, the actual efficiency of a heat engine will be, 1 - (T / T ) C H --------------- (6)

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##### ISBN: 9780130606204

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A clock pendulum oscillates at a frequency of 2.5 Hz. At t