Answer:
a = 7.35 ft / s²
Explanation:
For this exercise we must use the kinematics relations
x = v₀ t + ½ a t²
as the runner leaves the starting line his initial velocity is zero
x = ½ a t²
a = 
let's reduce the distance to foot
x = 60 yd (3ft / 1yd) = 180 ft
let's calculate
a = 2 180 / 7²
a = 7.35 ft / s²
Answer:
Where the electric potential is constant, the strength of the electric field is zero.
Explanation:
As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e
Eₓ = - dV / dx ----------(i)
From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.
<em>Therefore, a constant electric potential means that electric field is zero.</em>
