Asked by Justin
The driveshaft of a building’s air-handling fan is turned at 300 RPM by a belt running on a 0.3-m-diameter pulley. The net force applied by the belt on the pulley is 2000 N. determine the torque applied by the belt on the pulley, in N.m, and the power transmitted, in kW.
Answers
Answered by
bobpursley
torque= force*radius
power= torque*angular speed where
angular speed= 300*2PI/60 rad/sec
this will give you power in watts.
power= torque*angular speed where
angular speed= 300*2PI/60 rad/sec
this will give you power in watts.
Answered by
jackson
torque=300Nm
power=31.41 Watts
power=31.41 Watts
Answered by
Maddy
torque=F*D/2=2000n(0.15m)=300 N*m
Power = Torque*angular speed
=(300 N*m)(300 RPM) (2*PI/60)
=9424 W = 9.42 kW
These are the answers in the back of the text.
Power = Torque*angular speed
=(300 N*m)(300 RPM) (2*PI/60)
=9424 W = 9.42 kW
These are the answers in the back of the text.
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