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3 years ago

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Conductors and Insulators

1 answer:

grin007 [14]3 years ago

4 0

Answer:

<h2><em><u>Conductors</u></em></h2>

<em>The</em><em> </em><em>substances</em><em> </em><em>which</em><em> </em><em>allow</em><em> </em><em>an</em><em> </em><em>electric</em><em> </em><em>current </em><em>to</em><em> </em><em>pass</em><em> </em><em>through</em><em> </em><em>it</em><em> </em><em>is</em><em> </em><em>called</em><em> </em><em>conductors.</em>

<em>Example</em><em>:</em><em> </em><em>Metals</em><em> </em><em>like</em><em> </em><em>copper</em><em>,</em><em>iron</em><em> </em><em>,</em><em>graphite</em><em> </em><em>etc</em><em>.</em>

<h2><em><u>Insulators</u></em></h2>

<em>The</em><em> </em><em>substances</em><em> </em><em>which</em><em> </em><em>do</em><em> </em><em>not</em><em> </em><em>allow</em><em> </em><em>the</em><em> </em><em>electric</em><em> </em><em>current</em><em> </em><em>to</em><em> </em><em>pass</em><em> </em><em>through</em><em> </em><em>is</em><em> </em><em>called</em><em> </em><em>Insulators</em><em>.</em>

<em>Example</em><em>:</em><em> </em><em>Glass</em><em>,</em><em> </em><em>dry</em><em> </em><em>wood</em><em> </em><em>,</em><em>plastic</em><em> </em><em>etc</em><em>.</em>

<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em>

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Explanation:

<u>Rate of Change</u>

The volume of a cone of radius r and height h is given by

$\displaystyle V=\frac{\pi r^2h}{3}$

The height is said to be 1/2 of the radius, thus

$\displaystyle V=\frac{\pi r^2\cdot r}{2\cdot 3}$

$\displaystyle V=\frac{\pi r^3}{6}$

a) Knowing r=18 feet, the volume is

$\displaystyle V=\frac{\pi 18^3}{6}$

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b) The rate of change of the volume is computed by taking the derivative of both sides respect to the time

$\displaystyle V'=3\frac{\pi r^2}{6}r'$

$\displaystyle V'=\frac{\pi r^2}{2}r'$

Where r' is the given rate of change of the radius: 2 feet/day.

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Explanation:

Given that,

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Speed = 25.0 m/s

(a). We need to calculate the speed at the moment of being thrown

Using work energy theorem

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$-mg\times d=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

Put the value into the formula

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(b). We need to calculate the maximum height

Using work energy theorem

$[tex]W=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2$

$mg\times d=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2$

Here, $\dfrac{1}{2}mv_{2}^2$ =0

$-(mg)\times d=\dfrac{1}{2}mv_{1}^2$

$d=\dfrac{v_{1}^2}{2g}$

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Hence, (a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

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