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Power input = 700 WEfficiency = 75% so power output = 75% of 700 = 525 W

Time = 30 seconds => Work done = 525W * 30s = 15750 W.s = 15750 Joules = 15.75 kiloJoules.

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Q: An electric drill has an efficiency of 75 percent. If the motor has a power input of 700 watts and is run for 30 seconds. How much actual work is done. How would I get a kilojoules answer?
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