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nirvana33 [79]
3 years ago
13

A 20.0 μF capacitor initially charged to 30.0 μC is discharged through a 1.20 kΩ resistor. How long does it take to reduce the c

apacitor's charge to 15.0 μC
Physics
1 answer:
Kay [80]3 years ago
3 0

Answer:

16.6 ms or 0.0166 s

Explanation:

If Q is the final charge, Q' is the initial charge, C the capacitance ,R is the resistance , t the time taken  and τ the time constant,

[tex]Q = Q'( 1- e^{-t\div \tau })

τ = R C = (1.20×10³) (20×10⁻⁶) = 0.024 s

15 = 30 ( 1- e^{-t\div \ 0.024 })

( 1- e^{-t\div \ 0.024 }) = 15 ÷ 30

⇒ - e^{-t\div \ 0.024 }) = 0.5 -1  

⇒  e^{-t\div \ 0.024 }) = 0.5

Taking logarithm to the base e on both sides of this equation,

⇒ t = 0.0166 seconds = 16.6 milli seconds

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In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e
Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

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Thus,

\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_1-Q_3)}{Q_1}}

or

\eta=1-\frac{(Q_3)}{Q_1}}

also,

\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}

or

\frac{T_3}{T_1}=\frac{Q_3}{Q_1}

thus,

\eta=1-\frac{(T_3)}{T_1}}

thus,

\eta=1-\frac{(240\ K)}{500\ K}}

or

\eta=0.52

or

Efficiency = 52%

8 0
3 years ago
A force of 1 N will cause a mass of 1 kg to have an acceleration of 1 m/s2. Therefore, a force of 7 N applied to a mass of 7 kg
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1 m/s^2

Using F=ma,

7=7a

a=1

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(3) Dispersion forces result from the formation of instantaneous dipoles in a molecule or atom. When electrons are more concentrated in a place, instantaneous dipoles formed.

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Answer:

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