Posted: 2/22/2012 3:39:14 PM EDT
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I am pretty much self teaching because I started this class 5 weeks late. Not alot of help from the book for this one, thus my question. A correct answer would help, but I need the work (formula) so I can learn the process to be able to continue on.
An electric pump lifts 800N of water from a well 50 m deep in 90 seconds. The potential difference across the pump is 48.5 V, and a current of 10.3 A flows through the pump. a. How much work is done by the pump in lifting the water? My answer is 399640Nm.........correct? If so, no input needed. b. How much work is done on the pump by electrical current during the time interval? No FN idea........ |
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Quoted:
I am pretty much self teaching because I started this class 5 weeks late. Not alot of help from the book for this one, thus my question. A correct answer would help, but I need the work (formula) so I can learn the process to be able to continue on. An electric pump lifts 800N of water from a well 50 m deep in 90 seconds. The potential difference across the pump is 48.5 V, and a current of 10.3 A flows through the pump. a. How much work is done by the pump in lifting the water? My answer is 399640Nm.........correct? If so, no input needed. b. How much work is done on the pump by electrical current during the time interval? No FN idea........ a) 800 newtons x 50 meters = 40,000 NewtonMeters = 40KiloJoules b) 48.5 volts x 10.3 Amps x 90 seconds = 45kilowatt seconds = 45 KiloJoules c) efficiency of pump 40/45 = 89% |