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An electric motor drill, rated 350 W has an efficiency of 35%. The torque produced, if it is working at 3000 rpm. |
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Answer» 0.25 N m `omega = (2PI epsilon)/(60) = (2 pi xx 3000)/(60)` Let `tau` be the torque produced by the motor. Power produced = `tau omega = tau xx (2pi xx 3000)/(60)` Now `tau xx (2 pi xx 3000)/(60) = 35%` of 350 W or `tau xx 2pi xx 50 = (35)/(100) xx 350` or `tau = (35 xx 350)/(100 xx 2pi xx 50) = 0.39 Nm` |
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