<span> answer>>>>electric force <<<<by the way i don't like physics but i answer this for you ^-^</span>
<span>This is because centripetal force is just the net force of a circular motion. There are no attractive or repulsive forces here. This is not the case here. </span>
<span>The gravitational force is a force reliant on mass and attraction of the masses. There are attractive forces here, but not really repulsive forces. </span>
<span>The electric force is the only one that would make sense because it has to do with a relationship between charges and includes both repulsive and attractive forces.</span>
3. In a uniform electric field, the equation for the magnitude of the magnetic field is E=(V/d). V= voltage d= distance. If the magnetic field magnitude is
constant , as stated in your problem, then the voltage must stay the same otherwise the value of "E" would change". And the problem already told us the "E" is uniform and so, not changing. Does that make sense?
4a. If the magnetic field lines are equally spaced apart, in other words share the same
density. Then we know that the magnitude of the magnetic field is unchanging. This is because the density of of the magnetic field lines(how many are in a certain area) is related to the magnitude being expressed by the electric field. Greater magnitude is expressed by the presence of more lines (higher line density)
4b. The electric potential is measured in Volts(V) and is uniform along the same equipotential line. What is an equipotential line(gray)? It is a line drawn perpendicular(forms a right angle with) to the magnetic field lines(black) to show the changes in electric potential. One space where electric potential will always be the same because it will always be equal to 0 Volts is exactly in between a positive and negative charges of equal charge value I have pointed to this line with a purple arrow in my picture.
I really hope this makes sense to you and that my pictures help! :)
By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s
Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:
mass m = 0.170 kg
initial speed u = 6 m/s.
Distance covered s = 61 m
To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V
To do this, let us first calculate the kinetic energy at which the ball move.
K.E = 1/2m
K.E = 1/2 x 0.17 x 
K.E = 3.06 J
The work done on the ball is equal to the kinetic energy. That is,
W = K.E
But work done = Force x distance
F x S = K.E
F x 61 = 3.06
F = 3.06/61
F = 0.05 N
From here, we can calculate the acceleration of the ball from Newton second law
F = ma
0.05 = 0.17a
a = 0.05/0.17
a = 0.3 m/
To calculate the final velocity, let us use third equation of motion.
=
+ 2as
=
+ 2 x 0.3 x 61
= 36 + 36
= 72
V = 
V = 8.485 m/s
Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.
Learn more about dynamics here: brainly.com/question/402617
Answer:
The only parameter that changes is mass m
It is only necessary to calculate the ratio Eh/Ee

The kinetic energy of the heavy paricle is three times the kinetic energy of an electron