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3 years ago

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A circuit containing an inductor and a capacitor in series is designed to have a resonant frequency of 4511 Hz. If the inductor

has inductance 1.82 mH, then what is the required capacitance of the capacitor? Select one: a. 8.3e-4 Farads b.0.12 Farads c. 6.8e-10 Farads d. 2.7e-5 Farads e. 6.8e-7 Farads

1 answer:

OLEGan [10]3 years ago

3 0

Answer:

(e) $6.835\times 10^{-7}F$

Explanation:

At resonance we know that $X_l=X_C$

That is $\omega L=\frac{1}{\omega C}$

$\omega ^2=\frac{1}{LC}$

$\omega =\frac{1}{\sqrt{LC}}$

$f=\frac{1}{2\pi \sqrt{LC}}$

We have given resonance frequency f =4511 Hz and inductance L=1.82 mH

So $4511=\frac{1}{2\pi \sqrt{LC}}$

$LC=\frac{1}{4\pi ^2\times 4511^2}$

$LC=1.244\times 10^{-9}$

$C=\frac{1.244\times 10^{-9}}{1.82\times 10^{-3}}=0.6835\times 10^{-6}=6.835\times 10^{-7}F$

So option e is the correct answer

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Answer:

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Explanation:

The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:

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ΔT = 40°C - (-20°C) = 60°C

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How do I calculate the tension in the horizontal string?

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ANSWER

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Given:

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In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

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$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

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Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

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Answer:

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Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

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u= Free space permeability

I= Electrical current passing through the wire

a = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

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