Answer:Racquet force is twice of Player force
Explanation:
Given
ball arrives at a speed of 
ball returned with speed of 
average Force imparted by racquet on the ball is given by
where 
time of contact of ball with racquet
When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet
From 1 and 2 we get
Hence the magnitude of Force by racquet is twice the Force by player
If the rod is in rotational equilibrium, then the net torques acting on it is zero:
∑ τ = 0
Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:
• at the left end,
τ = + (50 N) (2.0 m) = 100 N•m
• at the right end,
τ = - (200 N) (5.0 m) = - 1000 N•m
• at a point a distance d to the right of the pivot point,
τ = + (300 N) d
Then
∑ τ = 100 N•m - 1000 N•m + (300 N) d = 0
⇒ (300 N) d = 1100 N•m
⇒ d ≈ 3.7 m
I think true. I'm pretty sure, but check w/ others too.
Answer:
Where is question 12, we need it to answer this question
Explanation:
