C. Gravitational force between two objects of masses m1/m2
is defined by Gm1xm2/r^2, where G is a universal constant, and r is the distance between the masses. This shows you why B is wrong; it would increase. Also, A is not a good choice, as shown. In the case of B, it would increase by say, 2/1 to 4/1 by doubling ones mass. In C's case the bottom would decrease on the fraction alot (square) if they were moved closer, thus the force would be greater.
At the top because it has the most potential to go down. meaning it is a greater distance
c. At the top
Answer:

Explanation:
Given that

As both charges are negative so there exist force of repulsion in direction as shown in figure.

Angle at which force F12 is acting is



Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction


Answer:
true
Explanation:
Reflection is when light bounces off an object. If the surface is smooth and shiny, like glass, water or polished metal, the light will reflect at the same angle as it hit the surface. ... This is called diffuse reflection. Diffuse reflection is when light hits an object and reflects in lots of different directions.
First, we will get the distance traveled before the driver applied the brakes.
distance = velocity * time
distance = 25*0.34 = 8.5 m
Now, we will calculated the distance that the car traveled after the driver applied the brakes. To do this, we will use the equation of motion:
<span>vf^2 = vi^2 + 2*a*d where:
</span>vf = zero, vi = 25 m/s and a = -7 m/s^2
Note: The negative sign is only to show deceleration
d = <span> 1/2*(625) /(7) = 44.6428 m
The total stopping distance =</span> 8.5 + 44.6428 = 53.1428 m