Question

1. An object is pivoted at a given point. A force of 6.0 N
applied at a normal distance of 25 cm from the pivoted point
rotates the given object. What is the torque of the force?​

Answer 1

Answer:

given,

F = 6.0 N

$r = 25 \: cm$

$= 0.25 \: m$

$torque = f \times r$

$= 6 \times 0.25$

$= 1.5Nm$