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scoundrel [369]
1 year ago
14

A set of four capacitors are attached to a 12V battery in the circuit shown below. All capacitances are measured in milli-Farads

. Find the amount of electric charge that resides on each capacitor once it is fully charged.

Physics
1 answer:
Bond [772]1 year ago
3 0

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

#SPJ1

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As we know that

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So, 45000000 = 45,000 kg

And,

1 kg = 0.001 tons  

So, 45000 kg = 45 tons    

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3 years ago
Samantha is checking the weather for her upcoming trip to Mexico City. The weather forecast predicts a high-pressure system for
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A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
bija089 [108]

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

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2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

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3 years ago
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We are given the mass of an <span>aluminum sculpture which is 145 kg and a horizontal force equal to 668 Newtons. The coefficient of friction can be determined by dividing the horizontal force by the weight of the object. In this case, 668 N / 145 * 9.8 equal to coeff of friction of 0.47</span>
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A current of 2.0A flows through a light bulb. What is the amount of charge flowing through the light bulb in 40 seconds?
lidiya [134]

Answer:

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Given the following data;

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Mathematically, the quantity of charge passing through a conductor is given by the formula;

Quantity of charge = current * time

Substituting into the formula, we have;

Quantity of charge = 2 * 40

Quantity of charge = 80 Coulombs

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