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Author Question: A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of ... (Read 95 times)

dbose

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on: Jul 29, 2018
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 2.00 rad/s2. What is the moment of inertia of the door about the hinges? A) 4.28 kgm2
  B) 7.52 kgm2
  C) 1.88 kgm2
  D) 0.684 kgm2
  E) 2.74 kgm2

Question 2

A 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of What is the angle (assumed to be less than 90) between the wrench handle and the direction of the applied force?
 
  A) 18
  B) 14
  C) 22
  D) 25


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amit

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Reply #1 on: Jul 29, 2018
Answer to Question 1

C

Answer to Question 2

A


dbose

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Reply #2 on: Jul 29, 2018
TY


amit

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Reply #3 on: Jul 29, 2018
You're welcome


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