he triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 103 N with an effective perpendicular lever arm of 2.80 cm, producing an angular acceleration of the forearm of 145 rad/s2. What is the moment of inertia of the boxer's forearm

Moment of inertia = 0.3862kg-m²

Explanation:

2.00x10³

2.80cm

145 rad

r = r⊥ x F

F is an applied force

r⊥ is the distance between the applied force and axis

Force exerted = 2.00x10³

r⊥ = 2.8cm = 0.028m

Alpha = 145rad/s²

r = 0.028m x 2.00x10³

r = 56.0N-m

To get the moment of inertia

56.0N-m² = (145rad/s²) x I

The I would be:

I = (56.0N-m²)/(145rad/s²)

I = 56/145

= 0.3862Kg-m²

This is the moment of inertia.

Thank you!

i think it would be 80 km/h because 30 + 50 = 80 (sorry if this not the right answer i'm trying my best)