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balandron [24]
2 years ago
15

If the index of refraction of a medium is 1.4, determine the speed of light in that medium.

Physics
1 answer:
Damm [24]2 years ago
3 0

The speed of light in that medium is 2.14 \times 10^8 \ m/s.

<u>Explanation:</u>

It is known that the light's speed is constant when it travels in vacuum and the value is 3 \times 108 m/s. When the light enters another medium other than vacuum, its speed get decreased as the light gets refracted by an angle.

The amount of refraction can be determined by the index of refraction or refractive index of the medium. The refraction index is measured as the ratios of speed of light in vacuum to that in the medium. It is represented as  η = \frac {c}{v}

So, here η is the index of refraction of a medium which is given as 1.4, c is the light's speed in vacuum (3 \times 10^8 ms^-^1) and v is the light's speed in that medium which we need to find.

1.4=  \frac{(3 \times 10 ^ 8)} {v}

v=  \frac {(3 \times 10^8)}{1.4} =2.14 \times 10^8 \ m/s

Thus the speed of light in that medium is 2.14 \times 10^8 \ m/s.

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A cylindrical wire has a length of 2.80 m and a radius of 1.03 mm. It carries a current of current of 1.35 A, when a a voltage o
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Answer:

0.023 Ohms

Explanation:

Given data

Length= 2.8m

radius= 1.03mm

current I= 1.35 A

voltage V= 0.032V

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V= IR

Now  R= V/I

Substitute

R= 0.032/1.35

R= 0.023 Ohms

Hence the resistance is 0.023 Ohms

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2 years ago
You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that
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Questions Diagram is attached below

Answer:

T=2.08s

Explanation:

From the question we are told that:

Speed of Train V=3.0m.s

Angle \theta=12\textdegree

Height of window h_w=0.90m

Width of window w_w=2.0m

The Horizontal distance between B and A from Trigonometric Laws is mathematically given by

 b=\frac{0.9}{tan12}

 b=4.23

Therefore

Distance from A-A

 d_a=2.0+4.23

 d_a=6.23

Therefore

Time Required to travel trough d is mathematically given as

 T=\frac{d_a}{v}

 T=\frac{6.23}{3}

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2 years ago
Part A A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the
kifflom [539]

Answer:

Height will be 3.8971 m        

Explanation:

We have given that radius of the solid r = 1.60 m

Mass of the solid disk m = 2.30 kg

Angular velocity \omega =4.46rad/sec

Moment of inertia is given by I=\frac{1}{2}mr^2

Transnational Kinetic energy is given by KE=\frac{1}{2}mv^2 as we know that v = v=\omega r

So KE=\frac{1}{2}m(\omega r)^2

Rotational kinetic energy is given by KE_{ROTATIONAL}=\frac{1}{2}I\omega ^2=\frac{1}{2}\left ( \frac{1}{2}mr^2 \right )\omega ^2=\frac{1}{4}m(r\omega )^2

Potential energy is given by mgh

According to energy conservation

mgh=\frac{1}{2}m(\omega r)^2+\frac{1}{4}m(\omega r)^2

h=\frac{3r^2\omega ^2}{4g}=\frac{3\times 1.60^2\times 4.46^2}{4\times 9.8}=3.8971m

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2 years ago
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