physics

What is the average useful power output of a person who does 5.50×106 J of useful work in 6.50 h?

Working at this rate, how long will it take this person to lift 1850 kg of bricks 1.20 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output here.)

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asked by stephanie
  1. assume you mean 5.50 * 10^6 Joules

    6.50 hr * 3600 s/hr = time in seconds

    power = Joules / time in seconds
    ( which is Watts )

    part B:
    power * time = 1850 * 9.81 * 1.20

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    posted by Damon

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