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3 years ago

A heat engine receives 7000 j of heat and loses 3000 j in each cycle. what is the efficiency of this engine?

Physics

1 answer:

Nonamiya [84]3 years ago

4 0

Answer: 57.1 %

Explanation:

The efficiency of the engine is given by:

$\eta=\frac{Q_{used}}{Q_{in}}$

where the numerator corresponds to the heat used by the engine, while the denominator corresponds to the heat in input to the engine.

In this problem, the heat in input is

$Q_{in}=7000 J$

while the heat used is

$Q_{used}=7000 J-3000 J=4000 J$

So, the efficiency is

$\eta=\frac{4000 J}{7000 J}=0.571=57.1\%$

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Answer:

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Explanation:

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3 years ago

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The question is in the picture

Sedbober [7]

Answer:

e) 120m/s

Explanation:

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which gives

$t_0 = \dfrac{2v_0}{g}.$

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$t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}$

$t_{tot} = \dfrac{3v_0}{g}.$

This $t_{tot}$ , we are told, is 36 seconds; therefore,

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$v_0 = \dfrac{36g}{3}$

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$\boxed{v_0 = 120m/s}$

which from the options given is choice e.

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3 years ago

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Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$