A 3.2 kg solid disk with a radius of 0.45 m has a tangential force of 420.4 N applied to it. What is the moment of inertia applied to the disk

Explanation:

Your question makes no sense.

moment of inertia is a property of the disk and its geometry.

The moment of inertia of a uniform solid disk around an axis through its geometric center and perpendicular to its flat ends is

I = ½mR² = ½(3.2)0.45² = 0.324 kg•m²

the applied torque about the same axis would be

τ = FR = 420.4(0.45) = 189.18 N•m

and the angular acceleration about the same axis would be

α = τ/I = 189.18/0.324 = 583.9 rad/s²

explanation:

newton's third law, for every action there is an equal and opposite reaction.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this