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DochEvi [55]
3 years ago
10

An empty parallel plate capacitor is connected between the terminals of a 4.4-V battery and charged up. The capacitor is then di

sconnected from the battery, and the spacing between the capacitor plates is quadrupled. As a result of this change, what is the new voltage between the plates of the capacitor?
I have:E = v/dE =9 Vthen the distance is doubled 9(2) = 18 V
Physics
1 answer:
torisob [31]3 years ago
7 0

Answer:

d

Explanation:

idk

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How much money would be saved by turning off one 100.0-W lightbulb 3.0 h/day for 365 days if the
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the money that would be saved is $13.14.

Explanation:

Given;

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3 0
2 years ago
You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?
Natali5045456 [20]
Equation of motion:

y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}

Since initial velocity is zero, the second term goes away:

y_{f}=y_{o}+0+ \frac{1}{2} at^{2}

y_{f}=y_{o}+\frac{1}{2} at^{2}

y_{f}-y_{o}= \frac{1}{2} at^{2}

y_{f}-y_{o}=5.4m

5.4m= \frac{1}{2} at^{2}

\frac{2(5.4m) }{a} = t^{2}

a = g = 9.81  \frac{m}{ s^{2}}

\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}

1.1 s^{2} = t^{2}\sqrt{1.1 s^{2}} =  \sqrt{t^{2}}

<u><em>t = 1.05s</em></u>
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because it is are of circle

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