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3 years ago

What is the formula for calculating the efficiency of a heat engine?

Physics

1 answer:

Olegator [25]3 years ago

7 0

Explanation:

The efficiency of a heat engine is defined as the ratio of the work output to the heat input into the system.

It is highly impossible to get an engine that is 100% efficient.
A heat engine that is 100% efficient suggests that all the heat input into the system is used to do work completely.
There is no heat loss.

Efficiency = $\frac{work output}{heat input}$ x 100

Efficiency has no unit

Learn more:

Heat brainly.com/question/914750

#learnwithBrainly

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(a) What is the tangential acceleration of a bug on the rim of a 13.0-in.-diameter disk if the disk accelerates uniformly from r

Oksi-84 [34.3K]

Answer:

0.321659377 m/s²

1.383138458 m/s

0.321659377 m/s²

0.62667 m/s²

0.7044 m/s² and 27.17°

Explanation:

d = Diameter of rim = 13 in = $13\times 0.0254=0.3302\ m$

r = Radius = $\frac{d}{2}=\frac{0.3302}{2}=0.1651\ m$

$\omega_f$ = Final angular velocity = $80\times\frac{2\pi}{60}=8.37758\ rad/s$

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 4.3 s

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{8.37758-0}{4.3}\\\Rightarrow \alpha=1.94827\ rad/s^2$

Tangential acceleration is given by

$a_t=r\alpha\\\Rightarrow a_t=0.1651\times 1.94827\\\Rightarrow a_t=0.321659377\ m/s^2$

The tangential acceleration of the bug is 0.321659377 m/s²

Tangential velocity is given by

$v=r\omega\\\Rightarrow v=0.1651\times 8.37758\\\Rightarrow v=1.383138458\ m/s$

The tangential velocity of the bug is 1.383138458 m/s

The tangential acceleration is constant which is 0.321659377 m/s²

Centripetal acceleration is given by

$a_c=\frac{a_tt^2}{r}\\\Rightarrow a_c=\frac{0.321659377^2\times 1}{0.1651}\\\Rightarrow a_c=0.62667\ m/s^2$

The centripetal acceleration of the bug is 0.62667 m/s²

The resultant of the acceleration gives us total acceleration

$a=\sqrt{a_t^2+a_c^2}\\\Rightarrow a=\sqrt{0.321659377^2+0.62667^2}\\\Rightarrow a=0.7044\ m/s^2$

Direction is given by

$\theta=cos^{-1}\frac{a_c}{a}\\\Rightarrow \theta=cos^{-1}\frac{0.62667}{0.7044}\\\Rightarrow \theta=27.17^{\circ}$

The magnitude and direction of the acceleration is 0.7044 m/s² and 27.17°

8 0

4 years ago

Please help! 50 points and brainliest to first correct answer!

sergiy2304 [10]

No. The companion is wrong.

You are shivering because, since it's cold outside, your body is becoming incapable of maintaining the the same temperature outside ad well as inside.

So, in order to compensate, your body uses mechanism of shivering in order to produce more heat by constricting the blood flow in vessels.

Therefore, you shiver from cold.

4 0

3 years ago

Read 2 more answers

In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright frin

atroni [7]

Answer:

Y = 0.0254 m = 25.4 mm

Explanation:

The formula for the fringe spacing in Young's Double-slit experiment is given by the following formula:

$Y = \frac{\lambda L}{d}$