Physics
Grade-9
Easy

Question

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

  1. 7229kJ
  2. 7229 J
  3. 7500 J
  4. 7400 J

hintHint:

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

The correct answer is: 7229 J


    • Efficiency (%) = (useful energy output ÷ energy input) × 100
    • E n e r g y space I n p u t equals fraction numerator u s e f u l space o u t p u t over denominator E f f i c i e n c y end fraction cross times 100 equals 6000 over 83 cross times 100 equals 7229 space J

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