A72.0-kg person pushes on a small doorknob with a force of 5.00 n perpendicular to the surface of the door. the doorknob is located 0.800 m from axis of the frictionless hinges of the door. the door begins to rotate with an angular acceleration of 2.00 rad/s 2 . what is the moment of inertia of the door about the hinges?

Moment of inertia will be

Explanation:

We have given mass of the person m = 72 kg

Radius r = 0.8 m

Force is given  F = 5 N

Angular acceleration

Torque is given by

We know that torque is also given by

, here I is moment of inertia and is angular acceleration

So

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a longitudinal wave moves energy that is to the direction of the wave’s movement. - true