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:
density of cube =11.605 g/cm³
Explanation:
density of a substance is the mass per unit volume of that substance.
the density of a substance = 
volume of a cube = l³,
l = 19.0mm , lets convert mm to cm
1mm = 0.1cm, thus, 19mm =19*0.1 =1.9cm
length of cube =1.9cm
volume of cube = 1.9³
density of cube = 
density of cube =11.605 g/cm³
Answer:
it would be letter E. near oceanic ridges
Explanation:
new ocean crust is formed at the mid ocean ridges
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:
m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,
m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂' --> equation 1
The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is
(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2
Solving equations 1 and 2 simultaneously, v₁' = -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
