We know, by conservation of energy :

Therefore,

Putting given values, we get :

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.
Hence, this is the required solution.
Answer:
25.33 rpm
Explanation:
M = 100 kg
m1 = 22 kg
m2 = 28 kg
m3 = 33 kg
r = 1.60 m
f = 20 rpm
Let the new angular speed in rpm is f'.
According to the law of conservation of angular momentum, when no external torque is applied, then the angular momentum of the system remains constant.
Initial angular momentum = final angular momentum
(1/2 x M x r^2 + m1 x r^2 + m2 x r^2 + m3 x r^2) x ω =
(1/2 x M x r^2 + m1 x r^2 + m3 x r^2 ) x ω'
(1/2 M + m1 + m2 + m3) x 2 x π x f = (1/2 M + m1 + m3) x 2 x π x f'
( 1/2 x 100 + 22 + 28 + 33) x 20 = (1/2 x 100 + 22 + 33) x f'
2660 = 105 x f'
f' = 25.33 rpm
Answer:

Explanation:
<u>Uniform Acceleration
</u>
When an object changes its velocity at the same rate, the acceleration is constant.
The relation between the initial and final speeds is:

Where:
vf = Final speed
vo = Initial speed
a = Constant acceleration
t = Elapsed time
It's known a train moves from rest (vo=0) to a speed of vf=25 m/s in t=30 seconds. It's required to calculate the acceleration.
Solving for a:

Substituting:


Add them together with south being negative. (-350 + 25) to get 325 south