
If the separation between the charges is increased then the magnitude of the force will increase in fact how the distance is being used in that formula.
1) Equivalent resistance, 1/R = 1/15 + 1/40 + 1/60 = 8+3+2 /120 = 13/120
R = 120/13 = 9.23
2) Current, I = V/R = 115/9.23 = 12.45 A
Answer:
There is an interval of 24.28s in which the rocket is above the ground.
Explanation:





From Kinematics, the position
as a function of time when the engine still works will be:

At what time the altitud will be
?
⇒ 
Using the quadratic formula:
.
How much time does it take for the rocket to touch the ground? No the function of position is:

Where our new initial position is
, the velocity when the engine breaks is
and the only acceleration comes from gravity (which points down).
Now, when the rocket tounches the ground:
Again, using the quadratic ecuation:

Now, the total time from the moment it takes off and the moment it tounches the ground will be:
.
Find the amount of work that the spring does. This can be found using the equation 1/2kx^2. Then, you must set that equal to the amount of kinetic energy the car has. This is possible thanks to the work-energy theorem.
1/2kx^2 = 1/2mv^2
Solve to find velocity. Remember, the spring is displaced .15 m, not 15!
To find the acceleration, use F = ma. The force being applied to the car is kx, and you know the mass. You do the math.
For problem C I don't know, haven't done that yet in my class. Sorry!