A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result of a directed force combined with a friction force. As a result of the applied torque the angular speed of the wheel increases from 0 to 10.3 rad/s. After 6.10 s the directed force is removed, and the wheel comes to rest 60.6 s later. (a) What is the wheel's moment of inertia (in kg m2)? kg m
(b) What is the magnitude of the torque caused by friction (in N m)? N m
(c) From the time the directed force is initially applied, how many revolutions does the wheel go through?
revolutions

Explanation:

= Torque = 36.5 Nm

= Initial angular velocity = 0

= Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

Torque is given by

The wheel's moment of inertia is

t = 60.6 s

= 10.3 rad/s

= 0

Frictional torque is given by

The magnitude of the torque caused by friction is

Speeding up

Slowing down

Total number of revolutions

The total number of revolutions the wheel goes through is .

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.

are there options? cause you ain’t. real specific