Physics

Grade-9

Easy

Question

# A television that is 83% efficient provides 6000 J of useful energy. How much energy does it consume?

- 7229kJ
- 7229 J
- 7500 J
- 7400 J

Hint:

- Efficiency (%) = (useful energy output ÷ energy input) × 100

## The correct answer is: 7229 J

- Efficiency (%) = (useful energy output ÷ energy input) × 100

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