. A shaft rotating at 3000rpm is transmitting a
power of 3.14KW. The magnitude of the
driving torque is
1) 6Nm 2) 10Nm 3) 15Nm 4) 22Nm
Answers
Answered by
6
Answer:
2) 10Nm
Explanation:
Given frequency ν =3000rpm=300060=50rps and
τ=3.14kW=3140w
the angular velocity ω=2πν=100π
Power P = torque×angular
speed=τ×ωor
Torque = τ =Pω=3140100π
=31.43.14=10 Nm.
hope...it...helps!
Answered by
2
Answer:
10Nm
Explanation:
Explanation:
Given angular velocity ν =3000rpm=300060=50rps
and
τ=3.14kW=3140ω
the angular velocity ω=2πν=100π
Power P = torque × angular speed = τ × ω
or
Torque = τ =Pω=3140100π =31.43.14=10 Nm.
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