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zalisa [80]
2 years ago
5

You take a couple of capacitors and connect them in series, to which you observe a total capacitance of 4.8microfarads. However,

when you connect them in parallel their combined capacitance is 35microfarads. Determine the value of each capacitor.
Physics
1 answer:
Naily [24]2 years ago
6 0

Answer:

C₁ = 34.8 μF

C₂ = 0.2 μF

Explanation:

When the capacitors are connected in series their resultant capacitance is given by the formula:

Cs = 1/C₁ + 1/C₂

where,

Cs = Series equivalent capacitance = 4.8 μF

C₁ = Capacitance of 1st Capacitor

C₂ = Capacitance of 2nd Capacitor

Therefore,

4.8 μF = 1/C₁ + 1/C₂

(4.8 μF)(C₁ C₂) = C₁ + C₂   --------------- equation 1

When the capacitors are connected in parallel their resultant capacitance is given by the formula:

Cp = C₁ + C₂

where,

Cp = Parallel equivalent capacitance = 35 μF

Therefore,

35 μF = C₁ + C₂   -------------- equation 2

solving equation 1 and equation 2 simultaneously, we get:

<u>C₁ = 34.8 μF</u>

<u>C₂ = 0.2 μF</u>

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8 0
2 years ago
At a baseball game, the batter hit a fly ball at time t = 0 s. The outfielder caught the ball at t = 5.8 s. When was the ball at
Agata [3.3K]
We have the following equation for height:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 Where,
 a: acceleration
 vo: initial speed
 h0: initial height.
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 For t = 0 we have:
 h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
 h (0) = h0
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 h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
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 vo = (1/2) * (9.8) * (5.8)
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 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
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 t = 28.42 / 9.8
 t = 2.9 seconds.
 Answer:
 
The ball was at maximum elevation when:
 
t = 2.9 seconds.
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