The JWST is postioned about 1.5 million kilometers from the earth on the side facing away from the sun
Answer:
109.32 N/m
Explanation:
Given that
Mass of the hung object, m = 8 kg
Period of oscillation of object, T = 1.7 s
Force constant, k = ?
Recall that the period of oscillation of a Simple Harmonic Motion is given as
T = 2π √(m/k), where
T = period of oscillation
m = mass of object and
k = force constant if the spring
Since we are looking for the force constant, if we make "k" the subject of the formula, we have
k = 4π²m / T², now we go ahead to substitute our given values from the question
k = (4 * π² * 8) / 1.7²
k = 315.91 / 2.89
k = 109.32 N/m
Therefore, the force constant of the spring is 109.32 N/m
Answer:
48.6°
Explanation:
The forward force, F equals the component of the weight along the slope.
So mgsinθ = ma where a = acceleration and θ = angle between the slope and the horizontal.
So a = gsinθ
Since we are given that a = 75%g = 0.75g,
0.75g = gsinθ
sinθ = 0.75
θ = sin⁻¹(0.75)
= 48.6°