Explanation:
It is given that,
Force applied to object, 
Position,
(b) The cross product of force and position vector is used to find the net torque about the z axis. It is given by :



or

The torque is acting in -z axis.
(a) The magnitude of torque is given by :


Hence, this is the required solution.
Answer:
Resultant is 152 N at 28.5 degrees south to the 100 N force
Explanation:
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Answer:
I = 8.75 kg m
Explanation:
This is a rotational movement exercise, let's start with kinetic energy
K = ½ I w²
They tell us that K = 330 J, let's find the angular velocity with kinematics
w² = w₀² + 2 α θ
as part of rest w₀ = 0
w = √ 2α θ
let's reduce the revolutions to the SI system
θ = 30.0 rev (2π rad / 1 rev) = 60π rad
let's calculate the angular velocity
w = √(2 0.200 60π)
w = 8.683 rad / s
we clear from the first equation
I = 2K / w²
let's calculate
I = 2 330 / 8,683²
I = 8.75 kg m