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IRINA_888 [86]
2 years ago
14

At t = 0 the switch S is closed with the capacitor uncharged. If C = 50 F,  = 20 V, and R = 4.0 k, what is the charge on the

capacitor when I = 2.0 mA?
Physics
1 answer:
gulaghasi [49]2 years ago
5 0

Hello!

Recall the equation for the current in an RC circuit as a capacitor is charging:
i(t) = \frac{\epsilon}{R}e^{-\frac{t}{RC}}

i(t) = Current as a function of time (2.0 mA)

ε = Emf of voltage source (20V)
R = Resistance of resistor (4kΩ)

C = Capacitance of capacitor (50 μF)

We can solve for the time at which the current in the circuit decreases to 2.0 mA by plugging in the given values and solving for 't'.

0.002 = \frac{20}{4000} e^{-\frac{t}{(4000)(0.00005)}}\\\\0.002 = 0.005 e^{-\frac{t}{(4000)(0.00005)}

Solve:
.4 = e^{-\frac{t}{0.2}}\\\\ln(.4) = -\frac{t}{0.2}\\\\t = -0.2(ln(.4)) = 0.1833 s

Now, we can solve for the charge on the capacitor using the equation for charge of a charging capacitor:
q(t) = C\epsilon(1 - e^{\frac{-t}{RC}})

Plug in the given values and the time we solved for.

q(t) = (0.00005)(20)(1 - e^{\frac{-0.1833}{(4000)(0.00005)}}) \\\\q(t) = 0.0006 = \boxed{600 \mu C}

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Answer:

\rho=995.50\ kg.m^{-3}

\bar w=9765.887\ N.m^{-3}

s=0.9955

Explanation:

Given:

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<u>So, mass of the empty can:</u>

m_c=\frac{w_c}{g}

m_c=\frac{0.153}{9.81}

m_c=0.015596\ kg

<u>Hence the mass of liquid(soda):</u>

m_l=m_f-m_c

m_l=0.369-0.015596

m_l=0.3534\ kg

<u>Therefore the density of liquid soda:</u>

\rho=\frac{m_l}{v_l} (as density is given as mass per unit volume of the substance)

\rho=\frac{0.3534}{3.55\times 10^{-4}}

\rho=995.50\ kg.m^{-3}

<u>Specific weight of the liquid soda:</u>

\bar w=\frac{m_l.g}{v_l}=\rho.g

\bar w=995.5\times 9.81

\bar w=9765.887\ N.m^{-3}

Specific gravity is the density of the substance to the density of water:

s=\frac{\rho}{\rho_w}

where:

\rho_w= density of water

s=\frac{995.5}{1000}

s=0.9955

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(b) The distance of mass from mass A if there is no gravitational force acted on C
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Answer:

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(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no  meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters

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m₁ = The mass of object 1

m₂ = The mass of object 2

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F₁₃ =  6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰

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0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0

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