7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the ca
pacitors?
1 answer:
Answer:
q = C V charge on 1 capacitor
q = 1 * 10E-6 * 110 = 1.1 * 10E-4 C per capacitor
N = Q / q = 1 / 1.1 * 10E-4 = 9091 capacitors
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(22-12)/4
10/4
2.5 meters per second sqaured
I think that numbers one, three, and four are true
C: if it senses unequal currents
50Kpa
Explanation:
P1V1 = P2V2
Where;
P1= 100. P2= ?
V1 = 500. V2 = 1000
100 × 500 = P2 × 1000
50000 = 1000P2
50000/1000 = P2
50 = P2
P2 = 50Kpa
A because it's the smaller the thicker you just can't have 0 gage
