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a 0.65 kg ball is moving in a circular path which has a radius of 1.35 meters with a linear speed of 4.5m/s. what is the centriperal force acting on this ball

Verified Answer :

The acceleration formula is
\(a_{c}=\frac{v^{2}}{r}\)

We are given
\(v=4.5\ m/s\)
\(r=1.35\ m\)
Plug in:
\(a_{c}=\frac{(4.5\ m/s)^{2}}{1.35\ m}=15\ m/s^{2}\)

The formula for force is
\(F=ma\)

We now have
\(a=15\ m/s^{2}\)
\(m=0.65\ kg\)
Plug in:
\(F=15\ m/s^{2}*0.65\ kg=9.75\ N\)

So the centripetal force is 9.75 Newtons.