If a heat engine takes in 4565 kJ and gives up 2955 kJ during one cycle, what is the engine’s efficiency?
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To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.
The expression for sound power is,
Here,
A = Area
I = Intensity
P = Power
At the same time the area can be written as,
Now the intensity is inversely proportional to the square of the distance from the source, then
The expression for the intensity at different distance is
Here,
= Intensity at distance 1
= Intensity at distance 2
= Distance 1 from light source
= Distance 2 from the light source
If we rearrange the expression to find the intensity at second position we have,
If we replace with our values at this equation we have,
Now using the equation to find the area we have that
Finally with the intensity and the area we can find the sound power, which is
Power is defined as the quantity of Energy per second, then