Answer:
1350N
Explanation:
[tex]2.75 ms = 2.75*10^{-3}s[/tex]
The force exerted on the hand would be the momentum divided by the duration of contact.
As the hand is coming to rest, final velocity would be 0
[tex]F = \frac{\Delta P}{\Delta t} = \frac{m(0 - v)}{\Delta t} = \frac{1.65*(2.25 - 0)}{2.75 * 10^{-3}} = -1350 N[/tex]
The magnitude of the force would be 1350N
