Answer with Explanation:
We are given that
Diameter=d=22.6 cm
Mass,m=426 g=
1 kg=1000 g
Radius,r=
1m=100 cm
Height,h=5m

a.By law of conservation of energy






Where 
b.Rotational kinetic energy=
Rotational kinetic energy=8.35 J
Answer:
A. It does not exhibit projectile motion and follows a straight path down the ramp.
Answer:
The maximum electrical force is
.
Explanation:
Given that,
Speed of cyclotron = 1200 km/s
Initially the two protons are having kinetic energy given by

When they come to the closest distance the total kinetic energy is converts into potential energy given by
Using conservation of energy


Put the value into the formula


We need to calculate the maximum electrical force
Using formula of force



Hence, The maximum electrical force is
.
Answer:

Explanation:
Given that,
The mass of a golf ball, m = 40 g = 0.04 kg
Its angular velocity, 
The radius of the sphere is 2.5 cm or 0.025 m
We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

Where I is moment of inertia
For sphere, 

So, the magnitude of the angular momentum of the sphere is
.