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andreev551 [17]
2 years ago
14

\textbf{Help with this question } " alt="\large \textbf{Help with this question } " align="absmiddle" class="latex-formula">~
\textsf{For an Isosceles prism of Angle A and} \textsf{Refractive index}\: \sf \mu ,\textsf{ it is found that }\textsf{the angle of minimum deviation }\sf{ \delta_m = A.}\textsf{Then which of the following is/are correct}

\textbf{[ Multiple choice ]}

\textsf{(a) At minimum deviation, the incident}\sf{ angle \: \:i_1 }\textsf{and the retracting angle} \sf{r_1}\textsf{ at the first retracting surface are} \textsf{related by }\sf{ r_1 = (i_1/2) }

\textsf{(b) For this prism, the retracting index}\sf{ \mu}\textsf{ and the angle of Prism A are related as}
\sf{ A = \dfrac{1}{2}cos^{-1}\bigg\{\dfrac{\mu}{2}\bigg\}}

\textsf{(c) For this prism, the emergent ray at the }\textsf{second surface will be tangential to the }\textsf{surface when the angle of incidence at the}\textsf{ first surface is}\: \sf{ i_1 = sin^{-1} \bigg[sinA \sqrt{4cos²\dfrac{A}{2}-1}-cosA\bigg]} [tex] \textsf{(d) For the angle of incidence}\sf{ i_1 = A,} \textsf{the ray}\textsf{ inside the prism is parallel to the base}\textsf{ of the prism}

Please provide explanation ~

​
Physics
1 answer:
PIT_PIT [208]2 years ago
8 0

<h3>Hi there !</h3><h2>Option A is correct </h2>

<h3> Please refer the attachment for explanation</h3><h2>Stay safe, stay healthy and blessed</h2><h2>Have a marvelous day</h2><h2>Thank you</h2>

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Answer:

F = 4.3671 * 10^{-8}\ Newtons

Explanation:

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F = GMm/d^2

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So we have that:

F = 6.67408*10^{-11} * 7.1*10^4 * 73/89^2

F = 4.3671 * 10^{-8}\ Newtons

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A capacitor is connected across an ac source. Suppose the frequency of the source is doubled. What happens to the capacitive rea
fgiga [73]

The capacitive reactance is reduced by a factor of 2.

<h3>Calculation:</h3>

We know the capacitive reactance is given as,

Xc = \frac{1}{2\pi fC}

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Xc\\ = capacitive reactance

f = frequency

C = capacitance

It is given that frequency is doubled, i.e.,

f' = 2f

To find,

Xc =?

Xc' = \frac{1}{2\pi f'C}

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Therefore, the capacitive reactance is reduced by a factor of 2.

I understand the question you are looking for is this:

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?

  1. The capacitive reactance is doubled.
  2. The capacitive reactance is traduced by a factor of 4.
  3. The capacitive reactance remains constant.
  4. The capacitive reactance is quadrupled.
  5. The capacitive reactance is reduced by a factor of 2.

Learn more about capacitive reactance here:

brainly.com/question/23427243

#SPJ4

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