Physics
Grade-9
Easy
Question
Your heating system is 55 percent energy-efficient. What amount of energy would it consume to transform 9000 kWh into useful thermal energy for heating the house during the winter?
- 16363.63 kWh
- 163.63 kWh
- 1.6363 kWh
- 1636 kWh
Hint:
Efficiency (%) = (useful energy output ÷ energy input) × 100
The correct answer is: 16363.63 kWh
- Efficiency (%) = (useful energy output ÷ energy input) × 100
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