2.57 joule energy lose in the bounce
.
<u>Explanation</u>:
when ball is the height of 1.37 m from the ground it has some gravitational potential energy with respect to hits the ground
Formula for gravitational potential energy given by
Potential Energy = mgh
Where
,
m = mass
g = acceleration due to gravity
h = height
Potential energy when ball hits the ground
m= 0.375 kg
h = 1.37 m
g = 9.8 m/s²

Potential Energy = 5.03 joule
Potential energy when ball bounces up again
h= 0.67 m

Potential Energy = 2.46 joule
Energy loss = 5.03 - 2.46 = 2.57 joule
2.57 joule energy lose in the bounce
Answer:
The acceleration of the satellite is 
Explanation:
The acceleration in a circular motion is defined as:
(1)
Where a is the centripetal acceleration, v the velocity and r is the radius.
The equation of the orbital velocity is defined as
(2)
Where r is the radius and T is the period
For this particular case, the radius will be the sum of the high of the satellite (
) and the Earth radius (
) :


Then, equation 2 can be used:
⇒ 


Finally equation 1 can be used:

Hence, the acceleration of the satellite is 
Answer:
2.1 rad/s
Explanation:
Given that,
Mass of a tether ball, m = 0.546 kg
Length of a rope, l = 4.56 m
The maximum tension the rope can withstand before breaking is 11.0 N
We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :

Let
is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

So, the maximum angular speed of the ball is 2.1 rad/s.
Answer:

Explanation:
given,
mass of wheel(M) = 3 Kg
radius(r) = 35 cm
revolution (ω_i)= 800 rev/s
mass (m)= 1.1 Kg
I_{wheel} = Mr²
when mass attached at the edge
I' = Mr² + mr²
using conservation of angular momentum





