Question

a motor with an efficiency of 75 percent must supply 240 j of useful work. what amount of work must be supplied to the motor

Answer 1

any one know it

Answer 2

Is a starting power factor of 0.36 and starting load of 87.3kW .... Gas engines have a very low load acceptance - often 10 - 20% for first step from .... factor on a motor with almost the exact same horsepower rating (60hp). ... There is also a plot of power factor vs time attached to the 1st post of that thread.

Answer 3

if 240 J is 75%, then: 240 + (1/3)240 = 100%

so 240 + 80 = 320 J must be supplied.

Answer 4

thank you aaronq

Answer 5

Another Question ok?