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:
His final velocity is 15.8 m/s.
Step-by-step explanation:
Given:
Initial velocity of the driver is,
m/s
Acceleration of the driver is,
m/s²
Time taken to reach final velocity is,
s.
The final velocity is given using the Newton's equations of motion as:
, where,
is the final velocity.
Now, plug in the given values and solve for
.

Therefore, his final velocity is 15.8 m/s.
The distance is 28 meters, and the displacement is -4.
For the distance it would be 12 + 16 = 28.
For the displacement it would be 12 - 16 = -4.
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Answer:
the time interval that an earth observer measures is 4 seconds
Explanation:
Given the data in the question;
speed of the spacecraft as it moves past the is 0.6 times the speed of light
we know that speed of light c = 3 × 10⁸ m/s
so speed of spacecraft v = 0.6 × c = 0.6c
time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds
Now, from time dilation;
t = Δt₀ / √( 1 - ( v² / c² ) )
t = Δt₀ / √( 1 - ( v/c )² )
we substitute
t = 3.2 / √( 1 - ( 0.6c / c )² )
t = 3.2 / √( 1 - ( 0.6 )² )
t = 3.2 / √( 1 - 0.36 )
t = 3.2 / √0.64
t = 3.2 / 0.8
t = 4 seconds
Therefore, the time interval that an earth observer measures is 4 seconds