ind the useful power output (in W) of an elevator motor that lifts a 2400 kg load a height of 35.0 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 2400 kg is raised in height, but the full 10,000 kg is accelerated. (Enter a number.). ? wWhat does it cost (in cents), if electricity is $0.0900 per kW · h? (Enter a number.)? cents (a) Find the useful power output (in W) of an elevator motor that lifts a 2400 kg load a height of 35.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of thecounterbalanced system is 10,000 kg-so that only 2400 kg is raised height, but the full 10,000 kg is accelerated. (Enter a number.)W(b) What does it cost (in cents), if electricity is $0.0900 per kW h? (Enter a number.)cents

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ind the useful power output (in W) of an elevator motor that lifts a 2400 kg load a height of 35.0 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 2400 kg is raised in height, but the full 10,000 kg is accelerated. (Enter a number.). ? w
What does it cost (in cents), if electricity is $0.0900 per kW · h? (Enter a number.)
? cents

(a) Find the useful power output (in W) of an elevator motor that lifts a 2400 kg load a height of 35.0 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 2400 kg is raised height, but the full 10,000 kg is accelerated. (Enter a number.)
W
(b) What does it cost (in cents), if electricity is $0.0900 per kW h? (Enter a number.)
cents

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