A 15-µF capacitor is charged to 40 V and then connected across an initially uncharged 25-µF capacitor. What is the final potential difference across the 25-µF capacitor?

24volts

Explanation:

If a 15-µF capacitor is charged to 40V, the charge across the capacitor can be calculated using the formula;

Q = CV where;

Q is the charge flowing across the capacitor

C is the capacitance of the capacitor. = 15-µF

V is the voltage = 40V

Q = 15×10^-6×40

Q = 0.0006coulombs

If the charge of 0.0006coulombs is then connected across an initially uncharged 25-µF capacitor, the potential difference across the 25-µF can be calculated using the initial expression;

Q = CV

V = Q/C

V = 0.0006/25×10^-6

V = 24Volts

Therefore, the final potential difference across the 25-µF capacitor will be 24volts

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr