A wheel of moment of inertia 10×10raise to power -3 kgm per second is making 40rev per second . the torque required to stop it in 20 s is ?
Answers
Answered by
0
Answer:
hwhehehehehjebejenenenenhebehebjenenennenebebebrbrbbrnrjr Kendrick yep Ricardo Ernesto stereo stereo two step awesome so spiritual artificial special provided h
Answered by
1
Explanation:
omega = omega not - alpha× time
0 = 40(2π) - alpha×20
alpha = 4π
torque = I alpha
= 4π × 10^-2
Similar questions