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Find the useful power output of an elevator motor that lifts a 2500 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 system is balanced by 10000 kg so that only 2500 kg is raised in height, but the full 10000 kg is accelerated.

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Sagar Yadav
Sagar YadavLv10
17 Nov 2020

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

 

  1. Find the useful power output of an elevator motor that lifts a 2500-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 2500 kg is raised in height, but the full 10,000 kg is accelerated. (b) What does it cost, if electricity is $0.0900 per kW h?

(a). Find the useful power output of an elevator motor that lifts a 2500-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 2500 kg is raised in height, but the full 10,000 kg is accelerated. (b) What does it cost, if electricity is  0.0900 $.

(a) Find the useful power output of an elevator motor that lifts a 2500-kg load at 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 2500 kg is raised in height, but the full 10,000 kg is accelerated.

(b) What does it cost, if electricity is $0.0900 per kW·h?

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