Physics, asked by Adityamagdum, 1 year ago

. 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 arenarohith
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 nvssarath
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.

I hope this was helpful.

Keep studying.

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