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
Answer:
The resulting velocity of the ball after it hits the racket was of V= 51.6 m/s
Explanation:
m= 55.6 g = 0.0556 kg
t= 2.8 ms = 2.8 * 10⁻³ s
F= 1290 N/ms * t - 330 N/ms² * t²
F= 1024.8 N
F*t= m * V
V= F*t/m
V= 51.6 m/s
the purpose of fuses and circuit breakers is (first answer)
Since force is mass*acceleration,
F = 70kg * 9.8 m/s
By Considering the vertical distance and both vertical and horizontal final velocity, the time t = 0.45 s and Velocity V = 6.7 m/s
Given that a Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table.
Height h = 1.0 m
As the ball rolls off the table, it will be fallen under gravity. Where
g = 9.8 m/
Initial vertical velocity
= 0
Initial horizontal velocity
= 5 m/s
Considering the vertical distance, the formula to use to calculate the time will be;
h = ut + 1/2g
1 = 0 + 1/2 x 9.8
1 = 4.9
= 1/4.9
t = 
t = 0.45 seconds
It takes 0.45 seconds to hit the floor if no one sneezes.
To calculate its velocity when it hits the floor, we will need to calculate for both vertical and horizontal final velocity and find the resultant velocity of the two.
Vertical component
=
+ gt
= 0 + 9.8(0.45)
= 4.41 m/s
Horizontal component
=
+ at
but a = 0
= 5 m/s
Final velocity V = 
V = 6.67 m/s
Therefore, it will hit the floor at a velocity of 6.7 m/s
Learn more here: brainly.com/question/5063616