Answer: Ok, first lest see out problem.
It says it's a Long cylindrical charge distribution, So you can ignore the border effects on the ends of the cylinder.
Also by the gauss law we know that E¨*2*pi*r*L = Q/ε0
where Q is the total charge inside our gaussian surface, that will be a cylinder of radius r and heaight L.
So Q= rho*volume= pi*r*r*L*rho
so replacing : E = (1/2)*r*rho/ε0
you may ask, ¿why dont use R on the solution?
since you are calculating the field inside the cylinder, and the charge density is uniform inside of it, you don't see the charge that is outside, and in your calculation actuali doesn't matter how much charge is outside your gaussian surface, so R does not have an effect on the calculation.
R would matter if in the problem they give you the total charge of the cylinder, so when you only have the charge of a smaller r radius cylinder, you will have a relation between r and R that describes how much charge density you are enclosing.
The battery provides electrical energy to run an electric circuit.
Avg sped = total distance/ total time = 1350 mi/ 5 hrs= 270mph (i dont know if ur teacher wants you to convert this to m/s)
300miles are traveled in 1 hr. So, 300 *2hrs = 600 miles south
750/250= 3hrs north
Total distance = 600 miles + 750 miles= 1350 miles
Total time is = 3hrs + 2hrs= 5hrs
Care este problema? Btw why are you speaking Romanian
