(b) 
We can actually solve first part (b) the problem. In fact, we know that the electric field strength at the surface of the a sphere is given by
where
k is the Coulomb's constant
Q is the charge on the surface of the sphere
R is the radius
For this sphere, the radius is half the diameter, so
We also know that the maximum charge is the value of charge deposited at which the electric field of the sphere becomes equal to the breakdown electric field, so
Solving the formula for Q, we find the maximum charge:
(a) 
The maximum potential of a charged sphere occurs at the surface of the sphere, and it is given by
where we already found at point b)
and we know that
R = 0.215 m
Solving for V, we find:
Answer:
a) net force = 110N to the right.
b) acceleration = 5.5 m/s/s
Explanation:
a) 150N - 40N = 110N
b) F = ma
110N = 20kg x acceleration
acceleration = 110/20 = 5.5
Answer:
m₂ = 1.2 [kg]
Explanation:
Since the explosion causes a force in a certain instant of time, for both bodies there is an equal impulse. In such a way that we can relate both impulses to be able to find the mass of the body needed in the next equation.
where:
m₁ = mass of the first piece = 0.75 [kg]
v₁ = velocity of the first piece = 7.2 [m/s]
m₂ = mass of the second piece [kg]
v₂ = velocity of the second piece = 4.5 [m/s]
![0.75*7.2=m_{2}*4.5\\m_{2}=1.2[kg]](https://tex.z-dn.net/?f=0.75%2A7.2%3Dm_%7B2%7D%2A4.5%5C%5Cm_%7B2%7D%3D1.2%5Bkg%5D)
Answer:
Explanation:
From the given question, the small sphere was provided with an excess charge of +3 C, while the smaller shell was given an excess of -7 C, it should be -7 C and not 7 C.
So, in light of that, to determine the electric charges values & signs on each of them, we have:
on a = +3 C
on b = -7 C
on c = -7 C
on d = +3 C
on e = -7 C
Um honestly I have no clue. Maybe you're in a cold room lol.
