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3 years ago

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Imagine a Carnot engine has a hot reservoir of 680 K and a cold reservoir of 220 K. What is the efficiency of the engine? 58.3%

75.9% 62.3% 72.1% 67.6%

2 answers:

Strike441 [17]3 years ago

7 0

N = 1 - (QC/QH)

n = efficiency

QC = Cold Reservoir = 220 K
QH = Hot Reservoir = 680 K

$n = 1 - (220 \div 680) (\div 10)\\ n = 1 - (22 \div 68) (\div 2) \\ n = 1 - (11 \div 34) \\ n = (34 - 11) \div 34 \\ n = 23 \div 34 \\ n = 0.676... \times 100\\ n = 67.6$

Answer: The engine's efficiency is 67.6% .

Romashka [77]3 years ago

6 0

the other person's right, it's E. 67.6% for plato users :)

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3 years ago

Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy

Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

$P_{1}V_{1}=nRT$

$V_{1}=\dfrac{nRT}{P_{1}}$

Put the value into the formula

$V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}$

$V_{1}=2.62\times10^{-3}\ m^3$

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

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3 years ago

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MA_775_DIABLO [31]

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From this simple formula, you can see that in order to relate a mass to a force, you need to know an acceleration. And if the acceleration changes, then the relationship between the force and the mass also changes. So there's no direct conversion.

ON EARTH ONLY, one kilogram of mass <em>weighs</em> 9.8 newtons. The acceleration that connects them is the acceleration of gravity on Earth. In other places, with different gravitational accelerations, 1 kilogram weighs more or less newtons.

But they don't convert directly. That would be like asking "How do you convert miles to miles-per-hour ?"

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3 years ago

Acceleration is defined as the rate of change for which characteristic?

Alenkasestr [34]

1) C. velocity

Acceleration is defined as the rate of change of velocity per unit time. In formulas:

$a=\frac{\Delta v}{\Delta t}$

where

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$\Delta t$ is the time interval

Therefore, the correct answer is C. velocity.

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Earth's gravity is a force, so it produces an acceleration on every object with mass located on the Earth's surface. This acceleration can be calculated, as it is given by the formula

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3 years ago

Read 2 more answers

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Citrus2011 [14]

Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

Let the speed of second asteroid is v.

The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

$\frac{GMm}{d^{2}}=\frac{mv^2}{d}$

$v=\sqrt{\frac{GM}{d}}$

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

$v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}$

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3 years ago