Answer:
um it well it will be 19.6 d) 9.8 J
Explanation:
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>
Answer:
Distance, d = 61.13 ft
Explanation:
It is given that,
Initial speed of the car, u = 50 mi/h = 73.34 ft/s
Finally, it stops i.e. v = 0
Deceleration of the car, 
We need to find the distance covered before the car comes to a stop. Let the distance is s. It can be calculated using third law of motion as :
s = 61.13 ft
So, the distance covered by the car before it comes to rest is 61.13 ft. Hence, this is the required solution.
Yes it is a force omg so smart
