**Answer:**

k = 1400.4 N / m

**Explanation:**

When the springs are oscillating a simple harmonic motion is created where the angular velocity is

w² = k / m

w =

where angular velocity, frequency and period are related

w = 2π f = 2π / T

we substitute

2π / T = \sqrt{ \frac{k}{m} }

T² = 4π²

k = π²

in this case the period is T = 1.14s, the combined mass of the children is

m = 92.2 kg and the constant of the two springs is

k = 4π² 92.2 / 1.14²

k = 2800.8 N / m

to find the constant of each spring let's use the equilibrium condition

F₁ + F₂ - W = 0

k x + k x = W

indicate that the compression of the two springs is the same, so we could replace these subtraction by another with an equivalent cosecant

(k + k) x = W

2k x = W

k_eq = 2k

k = k_eq / 2

k = 2800.8 / 2

k = 1400.4 N / m

Answer:

Option C. 30 m

Explanation:

From the graph given in the question above,

At t = 1 s,

The displacement of the car is 10 m

At t = 4 s

The displacement of the car is 40 m

Thus, we can simply calculate the displacement of the car between t = 1 and t = 4 by calculating the difference in the displacement at the various time. This is illustrated below:

Displacement at t = 1 s (d1) = 10 m

Displacement at t= 4 s (d2) = 40

Displacement between t = 1 and t = 4 (ΔD) =?

ΔD = d2 – d1

ΔD = 40 – 10

ΔD = 30 m.

Therefore, the displacement of the car between t = 1 and t = 4 is 30 m.

<span>The speed of longitudinal waves, S, in a thin rod = âšYoung modulus / density , where Y is in N/m^2.
So, S = âšYoung modulus/ density. Squaring both sides, we have, S^2 = Young Modulus/ density.
So, Young Modulus = S^2 * density; where S is the speed of the longitudinal wave.
Then Substiting into the eqn we have (5.1 *10^3)^2 * 2.7 * 10^3 = 26.01 * 10^6 * 2.7 *10^6 = 26.01 * 2.7 * 10^ (6+3) = 70.227 * 10 ^9</span>