The question is incomplete. The complete question is :
Two conducting spheres are mounted on insulating rods. They both carry some initial electric charge, and are far from any other charge. Their charges are measured. Then, the spheres are allowed to briefly touch, and the charge in one of them (sphere A) is measured again. These are the measured values:
a). Before contact:
Sphere A = 4.8 nC
Sphere B = 0 nC
What is the charge on sphere B after contact, in nC?
b). Before contact:
Sphere A = 2.9 nC
Sphere B = -4.4 nC
What is the charge on sphere B after contact, in nC?
Solution :
It is given that there are two spheres that are conducting and are mounted on an insulating rods which carry a initial charge and they are briefly touched and then one of the charge is measured.
Here the charge becomes divided when both the spheres are connected and then removed.
a). charge after they are charged


= 2.4 nC
b). The charge is


= -0.75 nC
Answer:
22.505 seconds
Explanation:
V =19.8m/s
V = a*to
t1 = 19.8/3.3
= 6seconds
Distance travelled during acceleration
= 1/2 x 3.3 x 6²
= 59.4m
X_total = x1 + x2
X2 = 373-59.4
X2 = 313.6m
t2 = x2/v
= 313.6/19.8
= 16.505
Total = 16.505 + 6
= 22.505 seconds
the minimum time in which an elevator can travel the 373 m from the ground floor is 22.505 seconds.
Answer:
The answer is 11N to the right
Explanation:
Because 4N-3N= 1N
Therefore, 12N-1N=11N
The netforce is 11N to the right, because the greatest force is 12N to the right so it is more likely that the object is being pulled to the right.
Answer:
The answer is A because the equation is KEi+PEi=KEf+PEf
i means initial (before) and f means final (after)