Answer:

Where
represent the force for each of the 5 cases
presented on the figure attached.
Explanation:
For this case the figure attached shows the illustration for the problem
We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.
Th formula is given by:

Where G is a constant 
represent the mass for the earth
represent the mass for the spaceship
represent the radius between the earth and the spaceship
For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.
Based on this case we can create the following rank:

Where
represent the force for each of the 5 cases
presented on the figure attached.
Answer:

Explanation:
<u>Displacement Vector</u>
Suppose an object is located at a position

and then moves at another position at

The displacement vector is directed from the first to the second position and can be found as

If the position is given as magnitude-angle data ( z , α), we can compute its rectangular components as


The question describes the situation where the initial point is the base of the mountain, where both components are zero

The final point is given as a 520 m distance and a 32-degree angle, so


The displacement is

Answer:
the tension is 18513N
Explanation:
Given that
mass = 1683kg
acceleration = 1.2m/s^2
acceleration due to gravity = 9.8m/s^2
T-mg = ma
T = ma + mg
T = m(a +g)
T = 1683 kg(1.20 m/s2 + 9.8)
T = 1683 (11)
T = 18513N
therefore, the tension is 18513N
The X and Y components are as follows;
1. X = 35 * cos 57 = 19. 1m/s; Y = 35 * sin 57 = 29.4 m/s
2. X = 12 * -cos 34 = -10 m/s; Y = 12 * -sin 34 = -6.7 m/s
3. X = 8 * -cos 90 = 0 m/s; Y = 12 -sin 90 = -8 m/s
4. X = 20 * cos 75 = 5. 2m/s; Y = 20 * (-sin 75) = -19.3 m/s
<h3>What are the horizontal and vertical components of the vectors?</h3>
The horizontal and vertical components of the velocities are given as follows:
- Horizontal component, X = x cos θ
- Vertical component, Y = y sin θ
1. 35 m/s at 57° from x-axis
X = 35 * cos 57 = 19. 1m/s
Y = 35 * sin 57 = 29.4 m/s
2. 12m/s at 34° S of W
X = 12 * -cos 34 = -10 m/s
Y = 12 * -sin 34 = -6.7 m/s
3. 8 m/s at South
X = 8 * -cos 90 = 0 m/s
Y = 12 -sin 90 = -8 m/s
4. 20 m/s at 275° from x-axis
X = 20 * cos 75 = 5. 2m/s
Y = 20 * (-sin 75) = -19.3 m/s
In conclusion, the X and Y components are found by taking cosines and sine of the angles.
Learn more about horizontal and vertical components at: brainly.com/question/26446720
#SPJ1
Answer:he's not applying force or motion
Explanation:
p=f & m