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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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A measurement has high<br> when it is very close to the<br> true value?
jarptica [38.1K]

Answer:

Accuracy

Explanation:

Accuracy means making measurements that are close to the value precision means making measurement that are close in value to eachother but not necessarily close to the true value.

I hope this helps! If not sorry.

7 0
3 years ago
For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with d
Sedaia [141]

Answer:

The maximum mass the bar can support without yielding = 32408.26 kg

Explanation:

Yield stress of the material (\sigma) = 200 M Pa

Diameter of the bar = 4.5 cm = 45 mm

We know that yield stress of the bar is given by the formula

                Yield Stress = \frac{Maximum load}{Area of the bar}

⇒                                \sigma = \frac{P_{max} }{A}  ---------------- (1)

⇒ Area of the bar (A) = \frac{\pi}{4} ×D^{2}

⇒                            A  = \frac{\pi}{4} × 45^{2}

⇒                            A = 1589.625 mm^{2}

Put all the values in equation (1) we get

⇒ P_{max} = 200 × 1589.625

⇒ P_{max} = 317925 N

In this bar the P_{max} is equal to the weight of the bar.

⇒ P_{max} = M_{max} × g

Where M_{max} is the maximum mass the bar can support.

⇒ M_{max} = \frac{P_{max} }{g}

Put all the values in the above formula we get

⇒ M_{max} = \frac{317925}{9.81}

⇒ M_{max} = 32408.26 Kg

There fore the maximum mass the bar can support without yielding = 32408.26 kg

3 0
3 years ago
CAN SOMEONE HELP ME PLEASE ASAP
Lady_Fox [76]
The third choice is correct
8 0
4 years ago
Read 2 more answers
How do force affect the acceleration of a body?<br><br>​
Arada [10]

The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

3 0
3 years ago
Helpp pleaseeeeee …..
Lilit [14]

Answer:

300 cos 30 = 40 a + 40 * .2 * 10

Total force = mass * acceleration + frictional force

260 = 40 a + 80

a = 180 / 40 = 4.5 m/s^2

Check:

15 a + 15 * 10 * .2 = T    acceleration of 15 kg block (assuming a = 4.5)

T = 15 (4.5) + 30 = 97.5     force required to accelerate 15 kg block

260 - 97.5 = 162.5     net force on 25 kg block

162.5 = 4.5 (25) + 25 * 10 * .2

162.5 = 112.5 + 50 = 162.5

4.5 m/s^2 checks out as correct

8 0
2 years ago
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