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Leviafan [203]
4 years ago
11

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]4 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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An aeroplane cruises at normal speed. Upon fl ying into turbulent (unstable) air, it suddenly drops downwards vertically. Passen
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Answer:

Your answer would be D. be pushed down into their seats.

Explanation:

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3 years ago
How does the input distance of a single fixed pulley compare to the out- put distance?
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3 years ago
An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the ce
Mamont248 [21]

Answer:

Explanation:

To find the angular velocity of the tank at which the bottom of the tank is exposed

From the information given:

At rest, the initial volume of the tank is:

V_i = \pi R^2 h_i --- (1)

where;

height h which is the height for the free surface in a rotating tank is expressed as:

h = \dfrac{\omega^2 r^2}{2g} + C

at the bottom surface of the tank;

r = 0, h = 0

∴

h = \dfrac{\omega^2 r^2}{2g} + C

0 = 0 + C

C = 0

Thus; the free surface height in a rotating tank is:

h=\dfrac{\omega^2 r^2}{2g} --- (2)

Now; the volume of the water when the tank is rotating is:

dV = 2π × r × h × dr

Taking the integral on both sides;

\int \limits ^{V_f}_{0} \ dV = \int \limits ^R_0 \times 2 \pi \times r \times h \ dr

replacing the value of h in equation (2); we have:

V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr

V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr

V_f = \dfrac{ \pi \omega ^2}{g} \Big [  \dfrac{r^4}{4} \Big]^R_0

V_f = \dfrac{ \pi \omega ^2}{g} \Big [  \dfrac{R^4}{4} \Big] --- (3)

Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then V_f  =  V_i

Replacing equation (1) and (3)

\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i

\omega^2 = \dfrac{4g \times h_i }{R^2}

\omega =\sqrt{ \dfrac{4g \times h_i }{R^2}}

\omega = \sqrt{\dfrac{4 \times 9.81 \ m/s^2 \times 0.7 \ m}{(0.5)^2} }

\omega = \sqrt{109.87 }

\mathbf{\omega = 10.48 \ rad/s}

Finally, the angular velocity of the tank at which the bottom of the tank is exposed  = 10.48 rad/s

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