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3 years ago

6

A 500.-kg roller coaster car starts from rest at the top of a 60.0-meter hill.

1 answer:

Paraphin [41]3 years ago

8 0

1.47x10^5 Joules
The gravitational potential energy will be the mass of the object, multiplied by the height upon which it can drop, multiplied by the local gravitational acceleration. And since it started at the top of a 60.0 meter hill, halfway will be at 30.0 meters. So
500 kg * 30.0 m * 9.8 m/s^2 = 147000 kg*m^2/s^ = 147000 Joules.
Using scientific notation and 3 significant figures gives 1.47x10^5 Joules.

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Answer:

The power output of this engine is $P = 17.5 W$

The the maximum (Carnot) efficiency is $\eta_c = 0.7424$

The actual efficiency of this engine is $\eta _a = 0.46$

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From the question we are told that

The temperature of the hot reservoir is $T_h = 1250 \ K$

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The energy exhausts into cold reservoir is $E_c = 7.4 *10^{4} J$

The power output is mathematically represented as

$P = \frac{W}{t}$

Where t is the time taken which we will assume to be 1 hour = 3600 s

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$W = E_h -E_c$

substituting values

$W = 63000 J$

So

$P = \frac{63000}{3600}$

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The Carnot efficiency is mathematically represented as

$\eta_c = 1 - \frac{T_c}{T_h}$

$\eta_c = 1 - \frac{322}{1250}$

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$\eta _a = \frac{W}{E_h}$

substituting values

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