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uranmaximum [27]
3 years ago
9

When two or more capacitors are connected in series across a potential difference:

Physics
1 answer:
34kurt3 years ago
7 0

Answer:

A) and B) are correct.

Explanation:

Let's take a look at the attached picture. Now

The total voltage across both capacitors is the same as the sum of the voltage from each device, that statement is true for any electrical device connected in series. So a) is TRUE

The equivalent capacitance is going to be: \frac{1}{C_{total}}=\frac{1}{C_1} +\frac{1}{C_2}

And that value can be mathematically proven that is always less than any of the values of each capacitor. So b is TRUE

And through both capacitors flow the same current, but the amount of charge depends on the value of the capacitors, so only could be the same if the capacitors are the same value. Otherwise, don't. C) not always, so FALSE

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(a) The work required to get the coaster to the top of the first hill is  1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

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P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

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2 years ago
A cylindrical water tank has a diameter of 9 ft and a height of 12 ft. the water surface is 2 ft from the top. about how much wa
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A weightlifter lifts a 1,250 N barbell 2 m in 3 s. How much power was used to lift the barbell?
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A capacitor consists of two square plates, 8.7 cm on a side, separated by a 2.0 mm air gap. How much energy would be stored in t
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Answer:

122.84 J

Explanation:

Since plate is square, area, A is given by (8.7/100)^{2}=0.007569m^{2}

The distance between plates, d, is given in the question as 2mm=0.002m

Charge on plate, Q, as given in the question is 240 \mu c

Assuming mica dielectric constant, k of 7

Capacitance, C is given by

C=\frac {k\epsilon_{o}A}{d}=\frac {(7)(8.85*10^{-12})(0.007569)}{0.002}=2.34*10*^{-10}F

Stored energy, E is given by

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3 years ago
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