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Westkost [7]
3 years ago
9

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
Physics
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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3 years ago
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Listed following is a set of statements describing individual stars or characteristics of stars. Match these to the appropriate
Oksana_A [137]

Answer:

Red giant or super giant → very cool but very luminous

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Explanation:

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6 0
3 years ago
A simple harmonic oscillator consists of a block (m = 0.50 kg) attached to a spring (k = 128 N/m). The block is pulled a certain
Zinaida [17]

Answer:

0.5 m

14.00595

8 m/s, 0.0625 s

5.71314 m/s

Explanation:

k = Spring constant = 128 N/m

A = Amplitude

E = Energy in spring = 16 J

Energy in spring is given by

E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m

The amplitude is 0.5 m

Time period is given by

T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s

Number of oscillations is given by

N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595

The number of oscillations is 14.00595

For maximum speed

\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s

The maximum speed is 8 m/s

For a distance of 0.5 m which is the amplitude

Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s

The time taken would be 0.0625 s

The maximum kinetic energy is equal to the mechanical energy

\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=16

At x = 0.35 m

v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s

The speed of the block is 5.71314 m/s

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