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bezimeni [28]
3 years ago
11

To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?

Physics
1 answer:
Nimfa-mama [501]3 years ago
8 0
The energy stored in a capacitor is given by:
U= \frac{1}{2}CV^2
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
C=3.0 \mu F = 3.0 \cdot 10^{-6} F
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V
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A weight suspended from a spring is seen to bob up and down over a total distance of 20 centimeters twice each second.
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<u>Answer</u>

8. 2 Hz

9. 0.5 seconds

10. 20 cm


<u>Explanation</u>

<u>Q 8</u>

Frequency is the number of oscillation in a unit time. It is the rate at which something repeats itself in a second.

In this case, the spring bob up and down 2 times per second.

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<u>Q 9</u>

Period is the time taken to complete one oscillation.

2 oscillations takes 1 second

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∴ Period = 0.5 seconds


<u>Q 10</u>

Amplitude is the the maximum displacement of the spring.

In this case the spring bob up 20 cm. This is it's displacement.

∴ Amplitude = 20 cm

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<h3>What is the condition for the electric force between two objects?</h3>

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brainly.com/question/22492496

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