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levacccp [35]
3 years ago
15

A heat engine: A) converts heat input to an equivalent amount of workB) converts work to an equivalent amount of heat C) takes h

eat in, does work, and loses energy heat D) uses positive work done on the system to transfer heat from a low temperature reservoir to a high temperature reservoir E) uses positive work done on the system to transfer heat from a high temperature reservoir to a low temperature reservoir
Physics
1 answer:
Masja [62]3 years ago
3 0

Answer:

C. Takes heat in, does work, and loses energy heat.

Explanation:

Heat engine is a system makes use of thermal energy (heat) to in order to do mechanical work.

This occurs by converting the heat into mechanical energy. This energy is then used to do work.

The key characteristic of a heat engine is that the substance with which work is done by, goes from a higher temperature to a lower temperature.

Hence, it loses heat as it does work.

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A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Serggg [28]

Answer:

The spring force constant is  k=243\ \frac{N}{m} .

Explanation:

We are told the mass of the ball is m=0.0555\ kg, the height above the spring where the ball is dropped is h=0.536\ m,  the length the ball compresses the spring is d=0.04897\ m and the acceleration of gravity is 9.8\ \frac{m}{s^{2}} .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy E_{mi} is equal to the final mechanical energy E_{mf} :

                                                    E_{mf}=E_{mi}

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

                                                   \frac{1}{2}kd^{2}=mgh

If we manipulate the equation we have that:

                                                    k=\frac{2mgh}{d^{2} }

                                         k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}

                                              k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}

                                                     k=243\ \frac{N}{m}

                                                   

                             

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