turning a doorknob through 0.25 of a revolution requires 0.14 J of work. What is the...

Question

Question

turning a doorknob through 0.25 of a revolution requires 0.14 J of work. What is the torque required to turn the doorknob?

science physics

Answer 1

Answer #1

We know that the torque is F * r, where r is the radius of the knob.

The work is F * 2 Pi* r * 0.25 = 0.14

=> F*r = 0.14 /(2*Pi * 0.25) =0.0891 Nm (answer)

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Answer 2

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