To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.
By Hook's law we know that force is defined as,

Where,
k = spring constant
x = Displacement change
PART A) For the case of the spring constant we can use the above equation and clear k so that




Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m
PART B) In the case of speed we can obtain it through the period, which is given by

Re-arrange to find \omega,



Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to




Therefore the mass of the trailer is 4093.55Kg
PART C) The frequency by definition is inversely to the period therefore



Therefore the frequency of the oscillation is 0.4672 Hz
PART D) The time it takes to make the route 10 times would be 10 times the period, that is



Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s
Answer:
F = 1.24*10^4 N
Explanation:
Given
Depth of the ship, h = 25 m
Density of water, ρ = 1.03*10^3 kg/m³
Diameter of the hatch, d = 0.25 m
Pressure of air, P(air) = 1 atm
Pressure of water =
P(w) = ρgh
P(w) = 1.03*10^3 * 9.8 * 25
P(w) = 2.52*10^5 N/m²
P(net) = P(w) + P(air) - P(air)
P(net) = P(w)
P(net) = 2.52*10^5 N/m²
Remember,
Pressure = Force / Area, so
Force = Area * Pressure
Area = πr² = πd²/4
Area = 3.142 * 0.25²/4
Area = 3.142 * 0.015625
Area = 0.0491 m²
Force = 0.0491 * 2.52*10^5
F = 12373 N
F = 1.24*10^4 N
We don't know Carter, and we don't know where he is or what
he's doing, so I'm taking a big chance speculating on an answer.
I'm going to say that if Carter is pretty much just standing there,
or, let's say, lying on the ground taking a nap, then the force of
the ground acting on him is precisely exactly equal to his weight.