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12345 [234]
3 years ago
7

Can someone help me with this please?

Physics
1 answer:
Slav-nsk [51]3 years ago
8 0

Answer:

C2, C1, C4, C5 and C6 are in parallel. Therefore, we use the formula Cp = C1 + C2 + ....

Cp = C2 + C1 + C4 + C5 + C6 = ( 7 * 10 ^-3) + (18 * 10^-6) + (0.8F) + (200 * 10^-3 F) + (750 * 10^-6) = 1.008F

Now, Cp will become one capacitor and it will be aligned with C3, therefore it will now become a circuit in series.

We use the formula: 1/Cs = 1/C1 + 1/C2 + .... + ....1/Cn

Thus,

1/Cs = 1/C3 + 1/Cp

1/Cs = 1/(14 * 10^-3 F) + 1/(1.008F)

Cs = 1.4 * 10 ^-2 or if we do not round too much it will give exactly 0.0138 F

So the answer should be a)

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We know that

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If 270 watts of power is used in 42 seconds, how much work was done<br>​
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Answer: W = 11340J

Explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

So this is the Formula: Power = Work / Time.

<u>Step 1:</u><em><u> Find the Formula</u></em>

P = W / T

<em><u> </u></em>

<u>Step 2: </u><u><em>Make W the subject of the equation.</em></u>

W = PT

<u>Step 3:</u><u> </u><u><em>Given.</em></u>

P = 270 Watts, T = 42 seconds

<u>Step 4:</u><u><em> Substitute these values into equation 2 .</em></u>

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<u>Step 5:</u><u> </u><u><em>Simplify.</em></u>

W = 11340J

The amount of work done was 11340.

~I hope I helped you! :)~

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