Answer:
In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion
In classical mechanics, the gravitational potential at a location is equal to the work (energy transferred) per unit mass that would be needed to move an object to that location from a fixed reference location. It is analogous to the electric potential with mass playing the role of charge. The reference location, where the potential is zero, is by convention infinitely far away from any mass, resulting in a negative potential at any finite distance.
In mathematics, the gravitational potential is also known as the Newtonian potential and is fundamental in the study of potential theory. It may also be used for solving the electrostatic and magnetostatic fields generated by uniformly charged or polarized ellipsoidal bodies
First we have to find out the gravity on that planet. We use Newton second equation of motion. It is given as,
s = ut +(gt^2)/2
Distance s = 25m
Time t = 5 s
Velocity u = 0
By putting these values,
25 = 1/2.g.(5)²
g = 2
So the gravity on that planet is 2. Lets find out the weight of the astronaut.
Mass of the astronaut on earth m = 80 kg
Weight of astronaut on earth W = mg = (80)(9.8) = 784 N
Weight of astronaut on earth like planet = (80)(2) = 160 N
x = 160N
The magnitude of that other charge will be 9×10⁻⁵ C. The force on the charge is inverse of the distance.
<h3>What is Columb's law?</h3>
The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The magnitude of that other charge is found as;
Hence, the magnitude of that other charge will be 9×10⁻⁵ C.
To learn more about Columb's law, refer to the link;
brainly.com/question/1616890
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Kinetic energy of the rock just before it hits the ground=KE=33000 J
Explanation:
Weight= 2200N
mg=2200
m(9.8)=2200
m=224.5 kg
initial velocity=0
final velocity =V
using kinematic equation V²=Vi²+2gh
V²=0+2 (9.8)(15)
V=17.1 m/s
now kinetic energy= 1/2 mV²
KE= 1/2 (224.5)(17.1)²
KE=33000 J
Thus the kinetic energy of the rock just before it hits the ground=33000 J
The real answer would be 16.5 but since you want to have a full number it would be 16
