Answer: 20 m/s
Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:
Ekinetic=(1/2)*(m*v^2)
then E=0.5*30Kg*(20 m/s)^2=15*400=6000J
6489 for the founding product
Answer:
The applied torque is 3.84 N-m.
Explanation:
Given that,
Moment of inertia of the wheel is
Initial speed of the wheel is 0 (at rest)
Final angular speed is 25 rad/s
Time, t = 13 s
The relation between moment of inertia and torque is given by :
So, the applied torque is 3.84 N-m.
Answer:
The Answer is D!
Explanation:
I checked it on Khan Academy.