Answer:

Explanation:
The angular momentum of an object is given by:

where
m is the mass of the object
v is its velocity
r is the distance of the object from axis of rotation
Here we have:
m = 350 g = 0.35 kg is the mass of the ball
v = 9.0 m/s is the velocity
r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)
Therefore, the angular momentum is:

Answer:
F=m(11.8m/s²)
For example, if m=10,000kg, F=118,000N.
Explanation:
There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

Solving for the force F, we obtain that:

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

In this case, the force from the rocket's thrusters is equal to 118,000N.
Answer:Expression given below
Explanation:
Given mass of spring
Compression in the spring
Let the spring constant be K
Using Energy conservation
potential energy stored in spring =Kinetic energy of Block


now conserving momentum


where
is the final velocity