An electric motor draws a current of 10 amp with a voltage of 110 V. The output shaft develops a torque of 10.2 N-m and a rotational speed of 1000 rpm. For steady-state operation, determine:.a) the electric power required by the motor and the power developed by the output shaft, each in kW. b) the net power input to the motor, in kW. c) the amount of energy transferred to the motor by electrical work and the amount of energy transferred out of the motor by the shaft.

Answer 1:

The input power into the mother is 1.1 kW

The output power at the shaft is = 1.068 kW

Explanation:

Electrical power = input = Volt*current = 10*110 = 1100W = 1.1kW

Motor shaft power = output = Torque*angular speed

Angular speed = 2¶N/60 = (2*3.142*1000)/60 = 104.4 rad/sec

Ouput = 10.2*104.4 =1068W = 1.068kW

Answer 2: net power input = 0.032kW

Explanation: net power input = power input-power output.

Answer 3: energy transfered into the motor = 1100J in one sec

Energy transfer to shaft is 1068J in one sec

Explanation: power is energy expended per unit sec.

b.transmitters send radio waves, and receivers capture radio waves.

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm