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2 years ago

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Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3600 Hz . To do this, the car a

larm circuitry must produce an alternating electric current of the same frequency. That's why your design includes an inductor and capacitor in series. The maximum voltage across the capacitor is to be 12.0 V . To produce a sufficiently loud sound, the capacitor must store an amount of energy equal to 1.55×10−2 J . What values of capacitance and inductance should you choose for your car-alarm circuit?

1 answer:

Alex17521 [72]2 years ago

6 0

Answer:

The capacitance and the inductance can choose for a car-alarm circuit are

C = 215.27 μF

L = 9.078 μH

Explanation:

$V =12.0 V$ , $E = 1.55*10^2 J$ , $f = 3600 Hz$

To determine the capacitance can use the equation

$U_c= \frac{1}{2}*C*V^2$

Solve to C'

$C = \frac{U_c*2}{V^2}=\frac{1.55x10^2J*2}{12.0^2V}$

$C=215.27 uF$

To find the inductance can use the frequency of the circuit

$f = \frac{1}{2\pi* \sqrt{C*L} }$

Solve to L'

$L = \frac{1}{4\pi^2*f^2*C}=\frac{1}{4\pi^2*3600^2*215.27 uF}}$

$L = 9.078 uH$

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Answer:

24.348mm

Explanation:

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K = d / €^n

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K = 345 / 0.02⁰.²² = 816mPa

The true strain based upon the stress of 414mPa =

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However the true relationship between true strain and length is given by

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2 years ago

The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan

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Answer:Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u> proportional to the current and <u><em>inversely</em></u> proportional to the distance from the wire. If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

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Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u> proportional to the current and <u><em>inversely</em></u> proportional to the distance from the wire.

Now if I'=3I and r'=2r then magnetic field B' is given by

$B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B$

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To answer the problem we would be using this formula which isv = sqrt(T/(m/L))
v = sqrt(100 N / [(0.100 kg)/(1.0 m)])
v = 31.62 m/s
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2 years ago

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To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

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The relation between the linear velocity and angular velocity is

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Where,

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At the same time we have that the centripetal acceleration is

$a_c = \frac{v^2}{r}$

$a_c = \frac{(r\omega)^2}{r}$

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2 years ago

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