The question is incomplete.
The distance between the Moon and Earth influences: 1) the attractive gravitational force between them, 2) the tides, 3) the eclipses, 4) the period of each full turn of the moon around the Earth.
Assuming the question refers to the gravitational attraction, we must use the fact that, as per, Newton's Universal Gravitaional Law, the attractive force between the two bodies is inversely related to the square distance that separates them.
Then, if the Moon were twice as far, the gravitational pull would be one fourth (1/4) of actual pull.
Answer:
1.92 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 200 Kg
Spring constant (K) = 10⁶ N/m
Workdone =?
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass (m) = 200 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = m × g
F = 200 × 9.8
F = 1960 N
Next we shall determine the extent to which the spring stretches. This can be obtained as follow:
Spring constant (K) = 10⁶ N/m
Force (F) = 1960 N
Extention (e) =?
F = Ke
1960 = 10⁶ × e
Divide both side by 10⁶
e = 1960 / 10⁶
e = 0.00196 m
Finally, we shall determine energy (Workdone) on the spring as follow:
Spring constant (K) = 10⁶ N/m
Extention (e) = 0.00196 m
Energy (E) =?
E = ½Ke²
E = ½ × 10⁶ × (0.00196)²
E = 1.92 J
Therefore, the Workdone on the spring is 1.92 J
To answer the following questions for this specific problem:
a. 11.48 secs
b. Vp = a*t*3.6 =
3*11.48*3.6 = 124.0 km/h
<span>c. 9.1 secs. </span>
I am hoping that this answer has satisfied your query about
and it will be able to help you.
Answer:
The motion is over-damped when λ^2 - w^2 > 0 or when 
> 0.86
The motion is critically when λ^2 - w^2 = 0 or when = 0.86
The motion is under-damped when λ^2 - w^2 < 0 or when < 0.86
Explanation:
Using the newton second law
k is the spring constante
b positive damping constant
m mass attached
x(t) is the displacement from the equilibrium position
Converting units of weights in units of mass (equation of motion)
From hook's law we can calculate the spring constant k
If we put m and k into the DE, we get
Denoting the constants
2λ = 
= 
λ = b/0.215
λ^2 - w^2 = 
This way,
The motion is over-damped when λ^2 - w^2 > 0 or when > 0.86
The motion is critically when λ^2 - w^2 = 0 or when = 0.86
The motion is under-damped when λ^2 - w^2 < 0 or when < 0.86
Answer:
33 N
Explanation:
v = Velocity of fluid = 8+2 = 10 m/s
= Density of fluid = 1.2 kg/m³
C = Coefficient of drag = 1.1
A = Cross sectional area = 0.5 m²
Drag force is given by
The drag force on the athlete is 33 N
