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Zepler [3.9K]
3 years ago
14

A parallel plate capacitor with air between the plates has a potential difference of 71.0 V. Determine the potential difference

across the plates when a dielectric material (κ = 4.50) is introduced, filling the space between the plates.... I attached the formulas needed to solve this btw

Physics
1 answer:
Gwar [14]3 years ago
6 0

Answer:

15.8 V

Explanation:

The relationship between capacitance and potential difference across a capacitor is:

q=CV

where

q is the charge stored on the capacitor

C is the capacitance

V is the potential difference

Here we call C and V the initial capacitance and potential difference across the capacitor, so that the initial charge stored is q.

Later, a dielectric material is inserted between the two plates, so the capacitance changes according to

C'=kC

where k is the dielectric constant of the material. As a result, the potential difference will change (V'). Since the charge stored by the capacitor remains constant,

q=C'V'

So we can combine the two equations:

CV=CV'\\CV=(kC)V'\\V'=\frac{V}{k}

and since we have

V = 71.0 V

k = 4.50

We find the new potential difference:

V'=\frac{71.0}{4.50}=15.8 V

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For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e
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Answer:

Explanation:

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ΔUs = 0

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ΔEth = negative .

b )

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= 400 x 2 = 800 J

energy output = 800 J

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It is 25% of metabolic energy output of his body

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= 4x 800 J .

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5 0
4 years ago
Assuming the diode is ideal but with a forward voltage drop of. 65 volts, what is the current in ma if v1=0v, and r1=490ω?
weeeeeb [17]

The current in the ideal diode with forward biased voltage drop of 65V is 132.6 mA.

To find the answer, we have to know more about the ideal diode.

<h3>What is an ideal diode?</h3>
  • A type of electronic component known as an ideal diode has two terminals, only permits the flow of current in one direction, and has less zero resistance in one direction and infinite resistance in another.
  • A semiconductor diode is the kind of diode that is used the most commonly.
  • It is a PN junction-containing crystalline semiconductor component that is wired to two electrical terminals.
<h3>How to find the current in ideal diode?</h3>
  • Here we have given with the values,

                       V_2=65V\\V_1=0V\\R_1=490Ohm.

  • We have the expression for current in mA of the ideal diode with forward biased voltage drop as,

                I=\frac{V_2-V_1}{R_1} =\frac{65}{490} =132.6mA

Thus, we can conclude that, the current in mA of the ideal diode with forward biased voltage drop of 65 V is 132.6.

Learn more about the ideal diode here:

brainly.com/question/14988926

#SPJ4

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2 years ago
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