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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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What do electric forces between charges depend on
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Answer:

On the magnitude of the charges, on their separation and on the sign of the charges

Explanation:

The magnitude of the electric force between two charges is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

From the formula, we see that the magnitude of the force depends on the following factors:

- magnitude of the two charges

- separation between the charges

Moreover, the direction of the force depends on the sign of the two charges. In fact:

- if the two charges have same sign, the force is repulsive

- if the two charges have opposite signs, the force is attractive

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Two protons are released from rest when they are 0.720 nm apart. For related problem-solving tips and strategies, you may want t
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Answer:

a) Speed of the electrons at maximum speed = (1.384 × 10⁴) m/s

The maximum speed occurs at the point where all of the initial potential energy is converted into kinetic energy.

b) Maximum acceleration of the protons = (2.660 × 10¹⁷) m/s²

The maximum acceleration occurs at the minimum distance apart for the two protons.

Explanation:

The maximum speed occurs when all the potential energy of the protons has been converted to kinetic energy.

The potential energy between the two protons at the instant of release is given by

U = (kq₁q₂/r)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

U = (kq²/r) = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ (7.2 × 10⁻¹⁰) = (3.204 × 10⁻¹⁹) N/m or Joules

At the maximum speeds, the two protons will not possess any potential Energy, only kinetic energy.

The sum of kinetic and potential energies is always constant for the system

(Initial Kinetic Energy) + (Initial Potential Energy) = (Kinetic Energy at maximum speed) + (Potential Energy at maximum speed)

Initial Kinetic Energy of the system = 0 J (Since both protons were intially at rest)

Initial Potential Energy = (3.204 × 10⁻¹⁹) J

Kinetic Energy at maximum speed = Sum of the kinetic energies of the protons at this point = (½mv²) + (½mv²) = (mv²) J (Since theu are both protons, they have the same mass and the same speed at maximum speed)

Potential Energy at maximum speed = 0 J

0 + (3.204 × 10⁻¹⁹) = mv² + 0

mv² = (3.204 × 10⁻¹⁹)

m = mass of a proton = (1.673 × 10⁻²⁷) kg

v = speed of each of the protons at maximum speed = ?

v = √[(3.204 × 10⁻¹⁹) ÷ m]

v = √[(3.204 × 10⁻¹⁹) ÷ (1.673 × 10⁻²⁷)]

v = √(1.915 × 10⁸) = 13,838.8 m/s = (1.384 × 10⁴) m/s

b) Since the two protons repel each other and force of repulsion reduces as the dI stance between the protons increases, the maximum acceleration occurs at the minimum distance apart for the two protons.

Force of repulsion acting on each proton is given through Coulomb's law as

F = (kq₁q₂/r²)

And the force acting on each proton is obtainable using Newton's law that

F = ma

So, the acceleration of each proton at any time is obtainable through a relation of these 2 formulas.

ma = (kq₁q₂/r²)

a = (kq₁q₂/r²m)

k = Coulomb' s constant = (8.988 × 10⁹) Nm²/C²

q₁ = q₂ = charge on a proton = q = (1.602 × 10⁻¹⁹) C

r = separation between the two protons = 0.72 nm = (7.2 × 10⁻¹⁰) m

m = mass of a proton = (1.673 × 10⁻²⁷) kg

a = [(8.988 × 10⁹) × (1.602 × 10⁻¹⁹)²] ÷ [(7.2 × 10⁻¹⁰)² × (1.673 × 10⁻²⁷)]

a = (2.660 × 10¹⁷) m/s²

Hope this Helps!!!

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kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

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Likurg_2 [28]

Answer:

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v = at + v₀

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t = 4 s

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