The correct field line would be A.
I'm not sure I completely understand the expression you want evaluated.
It looks like a fraction with the same exact thing in both the numerator and the denominator. A fraction like that always boils down to ' 1 '.
The rest energy of a particle is

where

is the rest mass of the particle and c is the speed of light.
The total energy of a relativistic particle is

where v is the speed of the particle.
We want the total energy of the particle to be twice its rest energy, so that

which means:


From which we find the ratio between the speed of the particle v and the speed of light c:

So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.
Answer:
im very con fused on what you mean by this
Explanation:
Answer:27 km per hour West + 17 km per hour North