Answer:
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The amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.
<h3>
Total capacitance of the circuit</h3>
The total capacitance of the circuit is calculated as follows;
Capacitors in series;
1/Ct = 1/8 + 1/7.5
1/Ct = 0.25833
Ct = 3.87 mF
Capacitors is parallel;
Ct = 3.87 mF + 12 mF + 15 mF
Ct = 30.87 mF
Ct = 0.03087 F
<h3>Charge in each capacitor</h3>
Q = CV
Q = 0.03087 x 12
Q = 0.37 C
Thus, the amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.
Learn more about capacitors here: brainly.com/question/13578522
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Answer: 704
Explanation:Vi = 0 m/s
vf = 65 m/s
a = 3 m/s2
d = ??
vf2 = vi2 + 2*a*d
(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d
4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d
(4225 m 2/m2)/(6 m/s2) = d
d = 704 m
Since there is no decimal point in the number given above, the counting for the number of the significant figures will start from the left. Then, the first zero from the left is insignificant. Therefore, in this number there are 6 significant figures.
