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3 years ago

A 6.00-μf parallel-plate capacitor has charges of 40.0 μc on its plates. how much potential energy is stored in this capacitor?

Physics

1 answer:

dedylja [7]3 years ago

7 0

Thew energy stored in a capacitor of capacitance $C$ and voltage between the plates $V$ is

$E=\frac{1}{2} CV^2=\frac{1}{2C} Q^2$ .

Substituting numerical value

$E=\frac{1}{2*6*10^{-6}} (40*10^{-6})^2\\ E=133.33\; \mu J$

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Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec

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Answer:

(a) $F_{net}$ = 4.19 x $10^{-5}$ N, and its direction is towards $m_{2}$ .

(b) It must be placed inside a hollow shell.

Explanation:

Let, $m_{1}$ = 165 kg, $m_{2}$ = 465 kg, $m_{3}$ = 60 kg, and the distance between $m_{1}$ and $m_{2}$ is 0.340 m.

(a) Since $m_{3}$ is placed midway between $m_{1}$ and $m_{2}$ , then its distance to both masses is 0.170 m.

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F = $\frac{Gm_{1}m_{2} }{r^{2} }$

Where all variables have their usual meaning.

Then,

a. $F_{net}$ = $F_{23}$ - $F_{13}$

$F_{13}$ = $\frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }$

= 2.25 x $10^{-5}$ N

$F_{23}$ = $\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }$

= 6.44 x $10^{-5}$ N

∴ $F_{net}$ = = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

The net force exerted by the two masses on the 60 kg object is 4.19 x $10^{-5}$ N.

(ii) / $F_{net}$ / = / $F_{23}$ / - / $F_{13}$ /

= 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

(iii) The direction of the net force is to the right i.e towards $m_{2}$ .

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

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3 years ago

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ad-work [718]

The answer is C 8.87*10^4 m/s (it shouldn't be m/s^2 though as velocity is in m/s)

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v = vo + at

When you plug in the values you get

v = 2.39*10^4 + 5400*12 . so v = 8.87*10^4 m/s. C is your answer.

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