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horsena [70]
3 years ago
8

A 610 W Carnot engine operates between constant-temperature reservoirs at 142°C and 69.9°C. What is the rate at which energy is

(a) taken in by the engine as heat and (b) exhausted by the engine as heat?
Physics
1 answer:
elixir [45]3 years ago
6 0

Answer:

rate at which energy taken is 3511.19 W

rate at which heat exhausted is 2901.19 W

Explanation:

Given data

power = 610 W

Temperatures T = 142°C  = 142 + 273 = 415 K

Temperatures T2 = 69.9°C  = 69.9 + 273 = 342.9 K

to find out

rate at which energy taken and heat exhausted

solution

we know the equation of efficiency of engine that is = 1 - (T2 / T1)

so efficiency = 1 - (342.9 / 415 )

efficiency is 0.17373

and we know efficiency = energy output / energy input

efficiency = energy output / energy input

0.17373 = 610 / energy input

energy input = 3511.19

so rate at which energy taken is 3511.19 W

and rate at which heat exhausted is 3511.19  - power

rate at which heat exhausted is 3511.19  - 610

rate at which heat exhausted is 2901.19 W

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A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.
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Answer:

a) The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

\vec p = m\cdot \vec v (1)

Where:

\vec p - Vector momentum, measured in kilogram-meters per second.

m - Mass of the particle, measured in kilograms.

\vec v - Vector velocity, measured in meters per second.

If we know that m = 2.93\,kg and \vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right], then the momentum is:

\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]

\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]

The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}} (2)

\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right) (3)

Where:

\|\vec p\| - Magnitude of momentum, measured in kilogram-meters per second.

\theta - Direction of momentum, measured in sexagesimal degrees.

If we know that p_{x} = 8.731\,\frac{kg\cdot m}{s} and p_{y} = -11.661\,\frac{kg\cdot m}{s}, then the magnitude and direction of momentum are, respectively:

\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}

\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}

\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)

\theta \approx 306.823^{\circ}

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

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