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
Newton's first and second laws of motion both do, but I think the one you're looking for is: <em>The First Law of Motion</em>. That description is a little more direct.
It says that if an object is not acted on by a net external force, then it continues in "constant, uniform motion".
We have the equation of motion
, where v i the final velocity, u is the initial velocity, a is the acceleration and s is the displacement
Here final velocity, v = 40m/s
Initial velocity, u = 0 m/s
Displacement s = 2 m
Substituting 
So the baseball pitcher accelerates at 400m/
to release a ball at 40 m/s.
To determine what the cyclists average speed is, simply divide the distance the cyclist has travelled by the time the cyclist has traveled for.
Assuming that this is the average rate the cyclist is moving at it would be 12 km/hr.
Answer:
The right hand rule states that, to find the direction of the magnetic force on a positive moving charge, the thumb of the right hand point in the direction of v, the fingers in the direction of B, and the force (F) is directed perpendicular to the right hand palm.