Velocity = displacement (distance)/time
v=80m/4s
v=20m/s
velocity = 20 meters per second
Answer:
the distance that the object is raised above its initial position is 5.625 m.
Explanation:
Given;
applied effort, E = 15 N
load lifted by the ideal pulley system, L = 16 N
distance moved by the effort, d₁ = 6 m
let the distance moved by the object = d₂
For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.
M.A = V.R

Therefore, the distance that the object is raised above its initial position is 5.625 m.
Answer:
a) K = 2/3 π G m ρ R₁³ / R₂
, b) U = - G m M / r
Explanation:
The law of universal gravitation is
F = G m M / r²
Part A
Let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / R₂
G m M / R₂² = m v² / R₂
v² = G M / R₂
They give us the density of the planet
ρ = M / V
V = 4/3 π R₁³
M = ρ V
M = ρ 4/3 π R₁³
v² = 4/3 π G ρ R₁³ / R₂
K = ½ m v²
K = ½ m (4/3 π G ρ R₁³ / R₂)
K = 2/3 π G m ρ R₁³ / R₂
Part B
Potential energy and strength are related
F = - dU / dr
∫ dU = - ∫ F. dr
The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1
U- U₀ = G m M ∫ dr / r²
U - U₀ = G m M (- r⁻¹)
We evaluate for
U - U₀ = -G m M (1 /
- 1 /
)
They indicate that for ri = ∞ U₀ = 0
U = - G m M / r
<span>B. a ship slowly sinking</span>
This is not balanced
To solve the problem it is necessary to use the concepts related to the calculation of periods by means of a spring constant.
We know that by Hooke's law

Where,
k = Spring constant
x = Displacement
Re-arrange to find k,




Perioricity in an elastic body is defined by

Where,
m = Mass
k = Spring constant


Therefore the period of the oscillations is 0.685s