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Hunter-Best [27]
3 years ago
6

When light in vacuum is incident at the polarizing angle (Brewster's angle) on a certain glass slab, the angle of refraction is

31.8 degree. Calculate the values of : i): the polarizing angle theta_p of the glass.
ii): the index of refraction, n, of the glass.
Physics
1 answer:
marysya [2.9K]3 years ago
3 0

Answer:

Explanation:

θ( p ) + θ( r ) = 90  

θ (r) = angle of refraction and θ ( p ) is polarising angle.

given θ ( r ) = 31.8

θ ( p ) = 90 - 31.8 = 58.2 degree

ii ) Tanθ ( p ) = n ( refractive index ) = Tan 58.2 = 1.6

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12345 [234]

Answer:

new moon

Explanation:

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A car slows down from 21 m/s to rest in a distance of 63m. Assuming the car has a constant acceleration, calculate the time it t
vladimir2022 [97]

Answer:

-3.5 m/s²

Explanation:

  • Initial Velocity = 21m /s
  • Final velocity = 0m/s
  • Distance = 63 m .
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\longrightarrow Stopping distance = u²/2(-a)

\longrightarrow 63m = (21m/s)² / -2a

\longrightarrow a = - 21 * 21 / 63 * 2 m/s²

\longrightarrow a = - 3.5 m/s²

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5 0
3 years ago
A gravitational field is generated by any object that has
WITCHER [35]

Answer:

two poles

Explanation:

5 0
3 years ago
Read 2 more answers
A 1.75 kg box is pushed with a 8.35 N force across ground where k = 0.267. What is the net force on the box?
lubasha [3.4K]

The net force acting on the box of mass 1.75kg when it is pushed across a ground with coefficient of friction k=0.267 with a force of 8.35N, is 3.77N.

To find the answer, we have to know more about the basic forces acting on a body.

<h3>How to find the net force on the box?</h3>
  • Let us draw the free body diagram of the given box with the data's given in the question.
  • From the diagram, we get,

                                 N=mg\\F_t=ma\\F_t=F-f

where, N is the normal reaction, mg is the weight of the box, F_t is the net force, f is the kinetic friction.

  • We have the expression for kinetic friction as,

                      f=kN=kmg=0.267*1.75*9.8= 4.58N

  • Thus, the net force will be,

                        F_t= 8.35-4.58=3.77N

Thus, we can conclude that, the net force acting on the box of mass 1.75kg when it is pushed across a ground with coefficient of friction k=0.267 with a force of 8.35N, is 3.77N.

Learn more about the basic forces on a body here:

brainly.com/question/28061293

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8 0
2 years ago
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A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg.
rodikova [14]

Answer:

the moment of inertia of the merry go round is 38.04 kg.m²

Explanation:

We are given;

Initial angular velocity; ω_1 = 37 rpm

Final angular velocity; ω_2 = 19 rpm

mass of child; m = 15.5 kg

distance from the centre; r = 1.55 m

Now, let the moment of inertia of the merry go round be I.

Using the principle of conservation of angular momentum, we have;

I_1 = I_2

Thus,

Iω_1 = I'ω_2

where I' is the moment of inertia of the merry go round and child which is given as I' = mr²

Thus,

I x 37 = ( I + mr²)19

37I = ( I + (15.5 x 1.55²))19

37I = 19I + 684.7125

37I - 19 I = 684.7125

18I = 684.7125

I = 684.7125/18

I = 38.04 kg.m²

Thus, the moment of inertia of the merry go round is 38.04 kg.m²

7 0
4 years ago
Read 2 more answers
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