efficiency = (useful energy transferred ÷ energy supplied) × 100
It's easy to use this formula, but we have to know both the useful energy and the energy supplied. The drawing doesn't tell us the useful energy, so we have to find a clever way to figure it out. I see two ways to do it:
<u>Way #1:</u>
We all know about the law of conservation of energy. So we know that the total energy coming out must be 250J, because that's how much energy is going in. The wasted energy is 75J, so the rest of the 250J must be the useful energy . . . (250J - 75J) = 175J useful energy.
(useful energy) / (energy supplied) = (175J) / (250J) = <em>70% efficiency</em>
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<u>Way #2: </u>
How much of the energy is wasted ? . . . 75J wasted
What percentage of the Input is that 75J ? . . . 75/250 = 30% wasted
30% of the input energy is wasted. That leaves the other <em>70%</em> to be useful energy.
The aluminum foil would reflect the most light !
The answer is D light rays shine on an object which then reflects back to our retina
Answer:
Car radiators: Water is used as coolant car radiators. Due to its high specific heat capacity, it can absorb a large amount of heat energy from the engine of the car, but its temperature does not rise too high.
Explanation:
i hope this answer your question if it s wrong please let know
If it is in its ground state, it will be in the level 1s
