Answer:
F = - K x
a) K = 1.3 kg * 9.8 m/s^2 / .096 m = 133 kg/sec^2
b) ω = (K/m)^1/2 angular frequency of SHM
ω = (133 / 1.3)^1/2 = 10.1 / sec
f = 2 π ω = 6.28 * 10.1 / sec = 63.5 / sec
P = 1/f = .0157 sec
Newton's 2nd law of motion:
Force = (mass) x (acceleration)
= (1,127 kg) x (6 m/s² forward)
= (1,127 x 6) newtons forward
= 6,762 newtons forward
______________________________
Momentum = (mass) x (speed)
= (69 kg) x (6 m/s)
= 414 kg-m/s
To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

Here,
M = Mass
R = Radius of the hoop
The precession frequency is given as

Here,
M = Mass
g= Acceleration due to gravity
d = Distance of center of mass from pivot
I = Moment of inertia
= Angular velocity
Replacing the value for moment of inertia


The value for our angular velocity is not in SI, then


Replacing our values we have that


The precession frequency is




Therefore the precession period is 5.4s
Answer:
9 times
Explanation:
Kinetic energy is:
KE = ½ mv²
When we triple the velocity, the kinetic energy increases by a factor of 9.
9KE = ½ m(3v)²