Answer:
a) 1.725*10^5 N
b) 3.83*10^3 N
c) i) 173.24 kN
c) ii) 4.57 kN
Explanation:
See the attachment for calculations
Answer:
the tangential velocity of the student is 4.89 m/s.
Explanation:
Given;
the radius of the circular path, r = 3.5 m
duration of the motion, t = 4.5 s
let the student's tangential velocity = v
The tangential velocity of the student is calculated as follows;

Therefore, the tangential velocity of the student is 4.89 m/s.
D). located out side the nucleus
Answer:
8.75
Explanation:
First, find the force of friction.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.9 m/s)² = F (1.4 m)
F = 11.7 N
Next, find the distance at the new velocity.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d
d = 8.75 m