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bagirrra123 [75]
3 years ago
15

If a heat engine takes in 4565 kJ and gives up 2955 kJ during one cycle, what is the engine’s efficiency?

Physics
1 answer:
vodomira [7]3 years ago
5 0

The engine efficiency is 64.73 %

<u>Explanation:</u>

Given data

To find the engine’s efficiency we have the formula,

Energy input- 4565 KJ

Energy output - 2955KJ

Efficiency= energy output/ energy input ×100%

 =2955/4565

  =0.6473 ×100

η               =64.73 %

The engine efficiency is 64.73 %

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<h2>Hello!</h2>

The answer is: B. Kinetic energy

<h2>Why?</h2>

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The gravity acceleration is equal to 9.81\frac{m}{s^{2}}, it means that when falling, the ball will increase it's speed 9.81m every second.

We can calculate the kinetic energy by using the following formula:

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Clean up the units first:

Cancel the C² on the right side, then divide each side by Newton:

8.926 x 10⁻³⁰ = (8.98755 x 10⁹ m²) x (1.602 x 10⁻¹⁹)² / D²

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Take the square root of each side:

<em>D = 5.08 meters</em>

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