(a) Find the useful power output (in W) of an elevator motor that lifts a 2300 kg load a height of 30.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg-so that only 2300 kg is raisedin height, but the full 10,000 kg is accelerated. (Enter a number.)63016.6(b) What does it cost (in cents), if electricity is $0.0900 per kW · h? (Enter a number.)cents

We know that the formula of the power is given as Power = EnergyTime We know that the energy is…

Find the useful power output (in W) of an elevator motor that lifts a 2300 kg load a height of 30.0 m in height, but the full 10,000 kg is accelerated. (Enter a number.) 63016.6 What does it cost (in cents), if electricity is s0.0900 per kw · h? (Enter a number.)

Find the useful power output (in W) of an elevator motor that lifts a 2300 kg load a height of 30.0 m in height, but the full 10,000 kg is accelerated. (Enter a number.) 63016.6 What does it cost (in cents), if electricity is s0.0900 per kw · h? (Enter a number.)

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter7: Work, Energy, And Energy Resources

Section: Chapter Questions

Problem 39PE: (a) Find the useful power output of an elevator motor that lifts a 2500-kg load a height of 35.0 m...

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Concept explainers

Instantaneous Power

Power is defined in physics as the amount of energy transmitted or transformed per unit time. The watt is the unit of electricity in the International System of Units, equivalent to one joule per second.

Question

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(a) Find the useful power output (in W) of an elevator motor that lifts a 2300 kg load a height of 30.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg-so that only 2300 kg is raised
in height, but the full 10,000 kg is accelerated. (Enter a number.)
63016.6
(b) What does it cost (in cents), if electricity is $0.0900 per kW · h? (Enter a number.)
cents

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