Kittykat5001 Kittykat5001

10-06-2021
Physics

contestada

An oven consumes 425 kWh of energy in order to provide 386 kWh of useful energy. What is its percent efficiency?

Respuesta :

warrenmoncaster35 warrenmoncaster35

10-06-2021

Answer:

90.8%

Explanation:

efficiency = output / input

Simply just put in what we know.

386/425 = 0.908 (Rounded too 3 s.f)

And if you want this in percentage, you simply x100 your answer. Hence the answer being 90.8%

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