Physics
Grade-9
Easy
Question
A computer that is 87% efficient consumes 475 kWh of energy. How much useful energy does it provide?
- 400 kWh
- 413.25 kWh
- 413.25 Wh
- 400 Wh
Hint:
- Efficiency (%) = (useful energy output ÷ energy input) × 100
The correct answer is: 413.25 kWh
- Efficiency (%) = (useful energy output ÷ energy input) × 100
- Useful Energy =
=
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