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Mila [183]
3 years ago
7

A heat engine operating between energy reservoirs at 20?c and 600?c has 30% of the maximum possible efficiency.

Physics
1 answer:
Anarel [89]3 years ago
8 0
 <span>The maximum possible efficiency, i.e the efficiency of a Carnot engine , is give by the ratio of the absolute temperatures of hot and cold reservoir. 
η_max = 1 - (T_c/T_h) 

For this engine: 
η_max = 1 - [ (20 +273)K/(600 + 273)K ] = 0.66 = 66% 

The actual efficiency of the engine is 30%, i.e. 
η = 0.3 ∙ 0.664 = 0.20 = 20 % 

On the other hand thermal efficiency is defined as the ratio of work done to the amount of heat absorbed from hot reservoir: 
η = W/Q_h 

So the heat required from hot reservoir is: 
Q_h = W/η = 1000J / 0.20 = 5000J</span>
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<span>The proton differs from the electron in sign although they have the same value. Like the electron, a proton will gain 215 electron-volts of eV in Kinetic energy. So 1.602Ă—10^-19 J * 215 = 344.43 * 10^(-19) J. But K. E. = mv^2 / 2, so v^2 = 2 * K.E/m. The mass of a proton is 1.673 * 10^-27 kg. So v = âš(2 * 344.43 * 10^(-19))/1.673Ă—10^-27 = 688.86 * 10^(-19)/1.673Ă—10^(-27) = 411.75 * 10^(-19-(-27)) = âš411.75 * 10^(8) = 202196.56 Also for the electron we have v^2 = 2 * K.E/m but here mass, m, = 9.109 * 10^-31 kg. So we have v = âš(2 * 344.43 * 10^(-19)) / 9.109 * 10^-31 = 688.86 * 10^(-19)/ 9.109 * 10^-31 = 75.624 * 10^(-19 - (-31)) = 75.624 * 10^(21) and v = 2.749 * 10^11</span>
3 0
3 years ago
Assuming the ball's initial velocity was 51 ∘ above the horizontal and ignoring air resistance, what did the initial speed of th
Umnica [9.8K]

horizontal distance of home run is 400 ft = 122 m

height of the home run is 3 ft = 0.9 m

now the angle of the hit is 51 degree

now we have equation of trajectory of the motion

x = vcos\theta * t

y = v sin\theta * t - \frac{1}{2} gt^2

solving above two equations we have

y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}

now here we will plug in all data

0.9 = 122 tan51 - \frac{9.8 * 122^2}{2*v^2 * cos^251}

0.9 = 150.65 - \frac{184150.2}{v^2}

\frac{184150.2}{v^2} = 149.75

v = 35.1 m/s

<em>so the ball was hit with speed 35.1 m/s from the ground</em>

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3 years ago
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Answer:

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