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vovikov84 [41]
3 years ago
12

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum. Find the refractive index of the medium.

Physics
2 answers:
nalin [4]3 years ago
8 0

Refractive index is defined as

\mu = \frac{speed of light in air}{speed of light in medium}

\mu = \frac{c}{v}

given that

v = 0.6 c

so we will have

\mu = \frac{c}{0.6c}

\mu = 1.67

grin007 [14]3 years ago
3 0
<h2>Answer:</h2>

The refractive index is 1.66

<h2>Explanation:</h2>

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum .

Refractive index of medium = speed of light in vacuum / speed of light in medium  

So

RI = 1/0.6 = 5/3 or 1.66

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if a stone is projected at an angle of 50 degrees to the horizontal with an initial velocity of 50m/s, what is the vertical comp
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Answer:

38.3 m/s

Explanation:

To find vertical component of initial velocity, you'd have to use sine ratio:

\displaystyle{\sin \theta = \dfrac{u_y}{u}}

\displaystyle{u_y} is vertical component of initial velocity and \displaystyle{u} is initial velocity given which is 50 m/s.

A stone is projected at an angle of 50 degrees so \displaystyle{\theta} = 50°. Substitute in the formula:

\displaystyle{\sin 50^{\circ} = \dfrac{u_y}{50}}\\\\\displaystyle{50 \sin 50^{\circ} = u_y}\\\\\displaystyle{u_y = 38.3 \ \, \sf{m/s}}

Therefore, the vertical component of initial velocity is approximately 38.3 m/s

(The picture is also attached for visual reference!)

3 0
2 years ago
Prove that the unit of area derived quantities are derived units​
frozen [14]

answer: derived physical quantities are those quantities that are obtained from the basic physical quantities by multiplication or division and area is one of them

8 0
3 years ago
Two planets, Dean and Sam, orbit the Sun. They each have with circular orbits, but orbit at different distances from the Sun. De
lyudmila [28]

Answer:

The correct answer is Dean has a period greater than San

Explanation:

Kepler's third law is an application of Newton's second law where the force is the universal force of attraction for circular orbits, where it is obtained.

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When applying this equation to our case, the planet with a greater orbit must have a greater period.

Consequently Dean must have a period greater than San which has the smallest orbit

The correct answer is Dean has a period greater than San

3 0
3 years ago
An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq
Andru [333]

Answer:

f = 409 Hz

Explanation:

We have,

Length of the open organ pipe, l = 0.29 m

Frequency of vibration of second overtone, f_2 = 1227 Hz

It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

f_2=\dfrac{3v}{2l}

v is speed of sound

Let f is the fundamental frequency. It is given by :

f=\dfrac{v}{2l}

The relation between f and f₂ can be written as :

f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz

So, the fundamental frequency of the pipe is 409 Hz.              

6 0
3 years ago
two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y
svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

v_0_x=6380 m/s

v_0_y=6770 m/s

time period of engine running=763 s

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y=7.27\times 10^6

Using s=ut+\frac{at^2}{2} in x and y direction

x=v_0_x\times t+\frac{at^2}{2}

4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}

4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}

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In y direction

y=v_0_y\times t+\frac{a't^2}{2}

7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}

7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}

a=7.24 m/s^2

x component=-1.23 m/s^2

y component=7.24 m/s^2

3 0
3 years ago
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