A6.00-mu f capacitor is connected to a 12.0-v battery. how much energy is stored in the capacitor? to the product of the capacitance of the capacitor times the potential difference across the capacitor. why isn't this equal to the stored energy? had the capacitor been connected to a 6.00-v battery, how much energy would have been stored? how does the energy stored in a capacitor depend on the potential difference across the capacitor?

Explanation:

C = 6 x 10^-6 F, V = 12 V,

Energy stored in the capacitor

E = 1/2 CV^2 = 0.5 x 6 x 10^-6 x 12 x 12 = 4.32 x 10^-4 J

Now the voltage is 6 V, So let energy be E'

E' = 1/2 CV^2 = 0.5 x 6 x 10^-6 x 6 x 6 = 1.08 x 10^-4 J

The energy stored in a capacitor is proportional to the square of teh potential difference applied across the capacitor.

i believe the answer is a hope it and is correct

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

where is unknown. because the position is parabolic, there is symmetry about the midpoint at , and we know that

so that the position function is

which means the velocity is

then the total distance traveled is

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at and ends at , so its displacement is 0.