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Kipish [7]
3 years ago
7

The angle of incidence of another red ray is 65º. The refractive index of the glass of block

Physics
1 answer:
Mamont248 [21]3 years ago
6 0

Answer:

34º

Explanation:

The computation of the angle of refraction in the glass for this ray is shown below:

Given that

the angle of incidence of another red ray is 65º

And, n_1 = 1

n_2 = 1.62

According to shell law

n_1 sin \theta_1 = n_2 sin \theta_r

sin \theta_r = 1 sin 65÷ 1.62

= 0.559

\theta_r  = sin^-1 (0.559)

= 34º

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A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c
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The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
  • Let's draw the free body diagram of the system using the given data.
  • From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
  • For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

                           N_x=86.62N

  • We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

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  • To find Ny, we need to find the tension T.
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                           F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N

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                    N_y= (40*9.8)-(169.8*sin59)=246.4N

  • Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

                 N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

Learn more about the tension here:

brainly.com/question/28106871

#SPJ1

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