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3 years ago

6

Camera flashes charge a capacitor to high voltage by switching the current through an inductor on and off rapidly. In what time

must the 0.100 A current through a 2.00 mH inductor be switched on or off to induce a 500 V emf

1 answer:

Ber [7]3 years ago

8 0

Answer:

The value is $\Delta t = 4.0 *10^{-7} \ s$

Explanation:

From the question we are told that

The current is $\Delta I = 0.100 \ A$

The inductor is $L = 2.0mH = 2.0*10^{-3} \ H$

The voltage induced is $\epsilon = 500 V$

Generally the emf induced is mathematically represented as

$\epsilon = L * \frac{\Delta I }{\Delta t }$

Here $\Delta t$ is the time taken

=> $\Delta t = \frac{L * \Delta I }{\epsilon }$

=> $\Delta t = \frac{2*10^{-3} * 0.100 }{500 }$

=> $\Delta t = 4.0 *10^{-7} \ s$

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Answer:

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Height of island $h=1.80\times10^{3}\ m$

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Initial velocity $v=2.55\times10^{2}\ m/s$

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We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

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Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

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Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

$t=37.03\ sec$

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

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Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

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