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3 years ago

A 9.12 microFarad capacitor is measured to have a reactance of 268.0 Ohm. At what frequency (in Hz) is it being driven?

Physics

1 answer:

Katen [24]3 years ago

3 0

Answer:

65.14 Hz

Explanation:

We have given the capacitance $C=9.12\mu F=9.12\times 106{-6}F$

And the capacitive reluctance = 268 ohm

The capacitive reluctance of a capacitive circuit is given by

$X_C=\frac{1}{\omega C}$

$X_C=\frac{1}{2\pi f C}$

$268=\frac{1}{2\pi f \times 9.12\times 10^{-6}}$

$f=\frac{1}{2\times 3.14\times 268\times 9.12\times 10^{-6}}=65.14Hz$

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3 years ago

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Answer:

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Explanation:

Given:

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3 years ago

A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin

Mashcka [7]

Answer:

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Explanation:

The general equation to calculate the center of mass is:

$r_{cm}=1/M*\int\limits {r} \, dm$

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$dm = \lambda*a*d\theta$ Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

$\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}$

So, the differential of mass is:

$dm = \frac{2M}{a\pi}*a*d\theta$

$dm = \frac{2M}{\pi}*d\theta$

Now we proceed to calculate X and Y coordinates of the center of mass separately:

$X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta$

$Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta$

Solving both integrals, we get:

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Therefore, the position of the center of mass is:

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3 years ago

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Answer:

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Explanation:

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3 years ago