Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
Protons
electrons are negatively charged
neutrons are neutral
Answer:
Kf= 36 J
W(net) = 32 J
Explanation:
Given that
m = 2 kg
F= 4 N
t= 2 s
Initial velocity ,u= 2 m/s
We know that rate of change of linear momentum is called force.
F= dP/dt
F.t = ΔP
ΔP = Pf - Pi
ΔP = m v - m u
v= Final velocity
By putting the values
4 x 2 = 2 ( v - 2)
8 = 2 ( v - 2)
4 = v - 2
v= 6 m/s
The final kinetic energy Kf
Kf= 1/2 m v²
Kf= 0.5 x 2 x 6²
Kf= 36 J
Initial kinetic energy Ki
Ki = 1/2 m u²
Ki= 0.5 x 2 x 2²
Ki = 4 J
We know that net work is equal to the change in kinetic energy
W(net) = Kf - Ki
W(net) = 36 - 4
W(net) = 32 J
