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Aliun [14]
2 years ago
9

A circuit contains four capacitors in parallel (10 F, 3 F, 7 F, and 1 F). What is the equivalent capacitance of this circuit?

Physics
1 answer:
Oxana [17]2 years ago
3 0

The equivalent capacitance (C_{eq}) of an electrical circuit containing four capacitors which are connected in parallel is equal to: A. 21 F.

<h3>The types of circuit.</h3>

Basically, the components of an electrical circuit can be connected or arranged in two forms and these are;

  • Series circuit
  • Parallel circuit

<h3>What is a parallel circuit?</h3>

A parallel circuit can be defined as an electrical circuit with the same potential difference (voltage) across its terminals. This ultimately implies that, the equivalent capacitance (C_{eq}) of two (2) capacitors which are connected in parallel is equal to the sum of the individual (each) capacitances.

Mathematically, the equivalent capacitance (C_{eq}) of an electrical circuit containing four capacitors which are connected in parallel is given by this formula:

Ceq = C₁ + C₂ + C₃ + C₄

Substituting the given parameters into the formula, we have;

Ceq = 10 F + 3 F + 7 F + 1 F

Equivalent capacitance, Ceq = 21 F.

Read more equivalent capacitance here: brainly.com/question/27548736

#SPJ1

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Alex pushes on a 2.0 kg book, resulting in a net force of 6.0 N on the book.
Yakvenalex [24]

Answer:

<h2>3.0 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a  = \frac{f}{m}  \\

From the question we have

a =  \frac{6}{2}  \\

We have the final answer as

<h3>3.0 m/s²</h3>

Hope this helps you

4 0
3 years ago
A gas in a cylinder expands from a volume of 0.110 m³ to 0.320 m³. heat flows into the gas just rapidly enough to keep the press
Elden [556K]

80000 Joule is the change in the internal energy of the gas.

<h3>In Thermodynamics, work done by the gas during expansion at constant pressure:</h3>

ΔW = -pdV

ΔW = -pd (V₂ -V₁)

ΔW = - 1.65×10⁵ pa (0.320m³ - 0.110m³)

     = - 0.35×10⁵ pa.m³

     = - 35000 (N/m³)(m³)

     = -35000 Nm

ΔW = -35000 Joule

Therefore, work done by the system = -35000 Joule

<h3>Change in the internal energy of the gas,</h3>

ΔV = ΔQ + ΔW

Given:

ΔQ = 1.15×10⁵ Joule

ΔW = -35000 Joule

ΔU = 1.15×10⁵ Joule - 35000 Joule

      = 80000 Joule.

Therefore, the change in the internal energy of the gas= 80000 Joule.

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brainly.com/question/14265296

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3 0
2 years ago
A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex
Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

\text{Efficiency of the heat engine} = \eta = 26\% = 0.26

\text{Energy taken in by the heat engine during each cycle} = Q_h

\text{Energy exhausted by the heat engine in each cycle} = Q_c = 8.55*10^3 J

\eta = 1 - \frac{Q_{c}}{Q_{h}}

0.26 = 1 - \frac{8.55\ast 10^{3}}{Q_{h}}

\frac{8.55* 10^{3}}{Q_{h}} = 0.74

Q_h = \frac{8.55*10^3}{0.74}

Q_h = 11.554*10^3J

PART A)

Work done by the heat engine in each cycle = W

W = Q_h-Q_c

W = 11.554*10^3J-8.55*10^3J

W = 3004J

According to the value given we have that,

P = 4.0kW

P = 4000W

Power is defined as the variation of energy as a function of time therefore,

P = \frac{W}{t}

4000W = \frac{3004J}{t}

t = \frac{3004}{4000}

t = 0.75s

Therefore the interval for each cycle is 0.75s

5 0
3 years ago
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