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3 years ago

The speed of light in a diamond is 1.22 x 108 m/s. Calculate the index of refraction for a diamond.

Physics

1 answer:

ololo11 [35]3 years ago

3 0

Answer:

The index of refraction of a diamond is 2.4175

Explanation:

refractive index

n = c/v

n = refractive index

c = speed of light

v = velocity of light thru the material

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Near the center of our galaxy, hydrogen gas is moving directly away from us in its orbit about a black hole. The electromagnetic

klemol [59]

To solve this problem it is necessary to apply the concept related to wavelength, specifically when the wavelength is observed from a source that is in motion to the observer.

By definition the wavelength is given defined by,

$\lambda_{obs} = \lambda_s \sqrt{\frac{1+u/c}{1-u/c}}$

Where

$\lambda_{obs}$ = Observed wavelength

$\lambda_s$ = Wavelength of the source

c = Speed of light in vacuum

u = Relative velocity of the source to the observer

According to our data we have that the wavelength emitted from the galaxy is 1875nm which is equal to the wavelength from the source, while the wavelength from the observer is $\lambda_{obs}=1945nm$

Therefore replacing in the previous equation we have,

$1945 = 1875 \sqrt{\frac{1+\frac{u}{c} }{1-\frac{u}{c} }}$

$\sqrt{\frac{1+u/c}{1-u/c}} = 1.03733$

$\frac{1+\frac{u}{c} }{1-\frac{u}{c} }=1.03733^2$

$1+\frac{u}{c} =1.03733^2*(1-\frac{u}{c} )$

Solving for u,

$1+\frac{u}{c} =1.03733^2*(1-\frac{u}{c} )$

$1+\frac{u}{c} =1.03733^2-1.03733^2(\frac{u}{c} )$

$\frac{u}{c} +1.03733^2\frac{u}{c} =1.03733^2-1$

$2.88595\frac{u}{c}=1.03733^2-1$

$\frac{u}{c} = \frac{1.03733^2-1}{2.88595}$

$u = \frac{1.03733^2-1}{2.88595}*c$

$u = 0.02635c$

Therefore the speed of the gas relative to earth is 0.02635 times the speed of light.

6 0

3 years ago

A ball rolls off a desk at a speed of 3 m/s and lands .40 seconds later. How far from the base of the desk does the ball land?

Salsk061 [2.6K]

Is the velocity constant? Is there any friction?

3 meters per second

then after 40 seconds it must 3*40 = 120 meters

120 meters or 0.12 km if you will

7 0

3 years ago

If you throw a ball down ward then acceleration immeditely after leaving your hand is

bezimeni [28]

Answer:

9.8m/s²

Explanation:

The acceleration of the ball thrown after leaving my hand is 9.8m/s². This will be the acceleration due to gravity on the body.

Acceleration due to gravity is caused by the pull of the earth on a massive object.
The value of this acceleration is 9.8m/s².
As the ball nears the surface, it comes near zero.

7 0

2 years ago

reasons. 5. Why is the unit of temperature called a fundamental unit? Give reasons. ring derived unit.

Furkat [3]

Explanation:

It doesn't depends upon other.

It have it's own identity.

It's a lot easier to measure temperature than to measure the motion of component particles.

8 0

2 years ago

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).

just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ( $\vec r_{cm}$ ), measured in meters, is defined by this weighted average:

$\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$\vec r_{i}$ - Location of the i-th particle with respect to origin, measured in meters.

If we know that $\vec r_{cm} = (-0.500\,m,-0.700\,m)$ , $m_{1} = 1\,kg$ , $\vec r_{1} = (-1.20\,m, 0.500\,m)$ , $m_{2} = 4.50\,kg$ , $\vec r_{2} = (0.600\,m, -0.750\,m)$ and $m_{3} = 4\,kg$ , then the coordinates of the third particle are: