Power Physics Example 3

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Example 3

easy
A 100 W100 \text{ W} light bulb is on for 2 hours2 \text{ hours}. How much energy does it use?

Solution

  1. 1
    Convert time: 2 h=7200 s2 \text{ h} = 7200 \text{ s}.
  2. 2
    Energy: E=Pt=100×7200=720,000 J=720 kJE = Pt = 100 \times 7200 = 720{,}000 \text{ J} = 720 \text{ kJ}

Answer

E=720 kJE = 720 \text{ kJ}
Energy equals power times time. A higher-wattage bulb uses more energy per second, leading to higher electricity costs.

About Power

The rate at which work is done or energy is transferred, measured in watts (joules per second).

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Example 2 medium

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Example 4 medium

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