## A 72.0-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

Question

A 72.0-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 0.52 rad/s2. What is the moment of inertia of the door about the hinges

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Physics
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2021-07-21T18:13:02+00:00
2021-07-21T18:13:02+00:00 1 Answers
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## Answers ( )

Answer:Explanation:We are given that

Mass,m=72 kg

Force,F=5 N

Distance,r=0.8 m

Angular acceleration,

We have to find the moment of inertia of the door about hinges.

We know that

Torque,

Moment of inertia,

Using the formula

Hence, the moment of inertia of the door about hinges=