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4 years ago

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A 20.0 F capacitor initially charged to 30.0 uC is discharged through a 3.00 k resistor. How long does it take to reduce the cap

acitor's charge to 15.0 C? Express your answer with the appropriate units. ?

1 answer:

Maslowich4 years ago

3 0

Answer:

t = 0.04159 s

Explanation:

when a capacitor and resister are connected in series. charge on its plates decreases exponentially

$q = qo * e^{\frac{-t}{RC}}$

where,

CHARGE q =15.0 C

INITIAL CHARGE q_0 = 30.0 uc

Resistor R = 20.0 F

capacitor C = 3.00

$15 = 30* e^{\frac{- t}{(20* 10^{-6} * 3* 1000)}$

t = 0.04159 s

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Answer:

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$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

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