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2 years ago

13

When Kevin pulls his cotton shirt off his body, the electrons get transferred from the (shirt or body) to the (shirt or body) .

So, the shirt becomes (positively or negatively) charged and Kevin’s body becomes (positively or negatively) charged.

1 answer:

Masja [62]2 years ago

6 0

<span>When Kevin pulls his cotton shirt off his body, the electrons get transferred from the shirt (in form of static charges i.e. electrons to the body. So, the shirt becomes positively charged and Kevin’s body becomes negatively charged.

As a result of charge transfer from the shirt to the body, we can hear a crackling sound. or if observed in dark, a sparkle can be seen.</span>

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Answer:

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Explanation:

Given that,

airliner cover distance s, in time t1 between the cities faces opposite A and B

Let the speed of the wind be v

and the speed of the plain be $v_p$

The speed of the plain is

$v_p = \frac{s}{t}$

Time taken from one side is

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Time taken during A and B is

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Time taken during B to A

$t_B_A = \frac{s/2}{s/t_1 -v}$

Total timetaken by plane at the presence of wind

$t = t_A_B + t_B_A$

$t = \frac{s/2}{\frac{s}{t_1} + v} + \frac{s/2}{\frac{s}{t_1} -v}$

rearrange the equation to get v

$v = \sqrt{(\frac{s}{t_1})^2-(\frac{s}{t_1t})^2 }$

Speed of the wind is $= \sqrt{(\frac{s}{t_1})^2-(\frac{s}{t_1t})^2 }$

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$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

$h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft$

Now y = 2500 ft

$cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974$

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