Saved Bookmarks
| 1. |
An electric motor of power 200 w is switched on for 1 minute and 40 seconds. If 60 percent of the energy of the motor is useful,calulate (a) useful work done by the motor is |
|
Answer» GIVEN, power of electric MOTOR , P = 200w motor is switched on for 1minute and 40 sec so, time , t = 60 + 40 = 100 sec. we know, Energy = power × time = 200 × 100 = 20000 J A/C to question, 60% of the energy of the motor is useful. so, useful energy = 60 % of 20000 = 60/100 × 20000 = 12000 J hence, useful workdone by the motor = 12000J |
|