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FromTheMoon [43]
3 years ago
6

The inductance in the drawing has a value of L = 9.4 mH. What is the resonant frequency f0 of this circuit?

Physics
1 answer:
yuradex [85]3 years ago
4 0

Answer:

The resonant frequency of this circuit is 1190.91 Hz.

Explanation:

Given that,

Inductance, L=9.4\ mH=9.4\times 10^{-3}\ H

Resistance, R = 150 ohms

Capacitance, C=1.9\ \mu F=1.9\times 10^{-6}\ C

At resonance, the capacitive reactance is equal to the inductive reactance such that,

X_C=X_L    

2\pi f_o L=\dfrac{1}{2\pi f_oC}

f is the resonant frequency of this circuit  

f_o=\dfrac{1}{2\pi \sqrt{LC}}

f_o=\dfrac{1}{2\pi \sqrt{9.4\times 10^{-3}\times 1.9\times 10^{-6}}}

f_o=1190.91\ Hz

So, the resonant frequency of this circuit is 1190.91 Hz. Hence, this is the required solution.

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Answer:

0.074 V

Explanation:

Parameters given:

Number of turns, N = 121

Radius of coil, r = 2.85 cm = 0.0285 m

Time interval, dt = 0.179 s

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Final magnetic field strength, Bfin = 97.9 mT = 0.0979 T

Change in magnetic field strength,

dB = Bfin - Bin

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dB = 0.0428 T

The magnitude of the average induced EMF in the coil is given as:

|Eavg| = |-N * A * dB/dt|

Where A is the area of the coil = pi * r² = 3.142 * 0.0285² = 0.00255 m²

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|Eavg| = |-121 * 0.00255 * (0.0428/0.179)|

|Eavg| = |-0.074| V

|Eavg| = 0.074 V

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The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a
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Let the the final kinetic energy of the canister be

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A 873-kg (1930-lb) dragster, starting from rest completes a 401.4-m (0.2509-mile) run in 4.945 s. If the car had a constant acce
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To solve this problem it is necessary to apply the kinematic equations of motion.

By definition we know that the position of a body is given by

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x_0 = Initial position

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a = Acceleration

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And the velocity can be expressed as,

v_f = v_0 + at

Where,

v_f = Final velocity

For our case we have that there is neither initial position nor initial velocity, then

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With our values we have x = 401.4m, t=4.945s, rearranging to find a,

a=\frac{x}{t^2}

a = \frac{ 401.4}{4.945^2}

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v_f = v_0 + at

v_f = 0 + (16.41)(4.945)

v_f = 81.14m/s

Therefore the final velocity is 81.14m/s

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