Pretty sure it’s A. Hope this helps.
That's called the "normal" to the surface at that point.
Answer:
(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East
(b) When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West
Explanation:
Given;
a force vector points due east,
= 140 N
let the second force = 
let the resultant of the two vectors = F
(a) When the resultant force is pointing along east line
the second force must be pointing due east


(b) When the resultant force is pointing along west line
the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.


Answer:
Range, 
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}