Answer: Option B. R = (1/2)gt^2
Explanation:
S = R (horizontal distance)
V^2 = 2gS
V^2 = 2gR
R = V^2 / 2g
But V = gt
R = (gt)^2 / 2g
R = (g^2 x t^2) / 2g
R = gt^2 / 2
But t^2 = 2h/g
R = ( g x 2h/g) / 2
R = h
But h = (1/2)gt^2
R = h = (1/2)gt^2
The planet that Punch should travel to in order to weigh 118 lb is Pentune.
<h3 /><h3 /><h3>The given parameters:</h3>
- Weight of Punch on Earth = 236 lb
- Desired weight = 118 lb
The mass of Punch will be constant in every planet;

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

where;
- M is the mass of Earth = 5.972 x 10²⁴ kg
- R is the Radius of Earth = 6,371 km
For Planet Tehar;

For planet Loput:

For planet Cremury:

For Planet Suven:

For Planet Pentune;

For Planet Rams;

The weight Punch on Each Planet at a constant mass is calculated as follows;

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.
<u>The </u><u>complete question</u><u> is below</u>:
Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.
Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).
<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>
Learn more about effect of gravity on weight here: brainly.com/question/3908593
6. Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.
F = ma
F = (15.3)(-9.8)
F = -150N
Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object. So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.
7. Same idea as question 2.
First determine the weight of the object:
F = ma
F = (30)(-9.8)
F = -294N
The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.
-294 + 150N + x = 0
x = 144N
So the person is exerting 144 N.
10. First find the force of block B to the right due to its acceleration:
F = ma
F = (24)(0.5)
F = 12N
So block B is moving 12N to the right relative to block A due to block A's movement to the left. However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A. The force that is causing block B to experience the lower relative force to the right is because of the friction. To find the friction:
The sum of the forces in the leftward and rightward direction for block B must equal 12N.
75 - x = 12
x = 63N
So the force of friction of block A on block B is 63N to the left.
Answer: car 1 is going how fast
Explanation: no need to answer without speed I won't know distance.
An independent variable is the variable that is changed or controlled in a scientific experiment to test the effects on the dependent variable. A dependent variable is the variable being tested and measured in a scientific experiment.