Is there more than one answer
The electric potential (V) produced by a point charge with a charge of magnitude Q, at a point a distance r away from the point charge, is given by the equation:

Here,
k = Coulomb's constant
Q = Charge
r = Distance
If we rearrange the equation to find the distance we have,



Now the value of the charge density is equivalent to the charge on the surface area of the sphere this is




Answer:
Combined speed, V = 2 m/s
Explanation:
Given that,
Mass of object 1, m = 5 kg
Speed of object 1, u = 4 m/s (due right)
Mass of object 2, m' = 2 kg
Speed of object 2, u' = -3 m/s (due left)
<em>Let V is the speed of the objects after the collision. Here, both objects stick to each other. Therefore, the momentum remains conserved. So,</em>

So, the combined speed of the masses will be 2 m/s.
Answer:
Since it is falling freely, the only force on it is its weight, w.
w = m × g = 250 kg × 9.8 m/s^2 = 2450 Newton/N