Answer:
The number density of the gas in container A is twice the number density of the gas in container B.
Explanation:
Here we have
P·V =n·R·T
n = P·V/(RT)
Therefore since V₁ = V₂ and T₁ = T₂
n₁ = P₁V₁/(RT₁)
n₂ = P₂V₂/(RT₂)
P₁ = 4 atm
P₂ = 2 atm
n₁ = 4V₁/(RT₁)
n₂ =2·V₁/(RT₁)
∴ n₁ = 2 × n₂
Therefore, the number of moles in container A is two times that in container B and the number density of the gas in container A is two times the number density in container B.
This can be shown based on the fact that the pressure of the container is due to the collision of the gas molecules on the walls of the container, with a kinetic energy that is dependent on temperature and mass, and since the temperature is constant, then the mass of container B is twice that of A and therefore, the number density of container A is twice that of B.
We make a graphic of this problem to define the angle.
The angle we can calculate through triangle relation, that is,

With this function we should only calculate the derivate in function of c

That is the rate of change of
.
b) At this point we need only make a substitution of 0 for c in the equation previously found.

Hence we have finally the rate of change when c=0.
Answer:
D. −F
Explanation:
the rest of the answers are
2/3F
The force is represented as a positive quantity and is repulsive.
Electrostatic force is inversely proportional to the square of the distance.
The direction of the force changes, and the magnitude of the force quadruples.
hope this helps sorry if i was too late! :)