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:
0.025 m
Explanation:
From the question,
Applying Hook's law
F = ke................... Equation 1
Where F = Force, k = spring constant of the scale, e = maximum distance at which the spring will compress.
make e the subject of the equation
e = F/k....................... Equation 2
Given: F = 10 N, e = 395 N/m
Substitute these values into equation 2
e = 10/395
e = 0.025 m
D.
50 mph - 30 mph= 20 mph net velocity
change.
20mph/3600 seconds/hour= .00555 MPS
.0055 miles per second
40 seconds to complete the change
.0055/40= .000138
Answer:
98 m √
Explanation:
How about s = Vo * t + ½at² ?
s = h = Vo * 2s - 4.9m/s² * (2s)² = 2Vo - 19.6
and
h = Vo * 10s - 4.9m/s² * (10s)² = 10Vo - 490
Subtract 2nd from first:
0 = -8Vo + 470.4
Vo = 58.8 m/s
h = 58.8m/s * 2s - 4.9m/s² * (2s)² = 98 m