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Tanya [424]
2 years ago
15

Define refractive index.​

Physics
2 answers:
levacccp [35]2 years ago
8 0
The refractive index of a material is a dimensionless number that describes how fast light travels through the material. It is defined as n={\frac {c}{v}}, where c is the speed of light in vacuum and v is the phase velocity of light in the medium.

the ratio of the velocity of light in a vacuum to its velocity in a specified medium.
irga5000 [103]2 years ago
5 0

Answer:

In optics, the refractive index of a material is a dimensionless number that describes how fast light travels through the material. It is defined as n={\frac {c}{v}}, where c is the speed of light in vacuum and v is the phase velocity of light in the medium.

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How does gas exert pressure? Explain using the change in momentum.
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Verify that the linear speed of an ultracentrifuge is about 0.50 km's, and Earth in its orbit is about 30 km/s by calculating:
FrozenT [24]

Answer:

a) Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

Explanation:

The linear speed of the particle (v), measured in kilometers per second, rotating in a circular pattern is calculated by the following formula:

v = R\cdot \omega (1)

Where:

R - Radius, measured in kilometers.

\omega - Angular speed, measured in radians per second.

Now we proceed to calculate the linear speed of each element:

a) Ultracentrifuge

If we know that \omega \approx 5235.988\,\frac{rad}{s} and R = 1\times 10^{-4}\,km, then the linear velocity is:

v = (1\times 10^{-4}\,km)\cdot \left(5235.988\,\frac{rad}{s} \right)

v = 0.524\,\frac{km}{s}

Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Earth

The Earth is 150 million kilometers away from the Sun and takes 365 days to complete one revolution around the Sun. First, we calculate angular speed of the planet:

\omega = \frac{2\pi}{T} (2)

Where T is the period, measured in seconds.

If we know that T = 31536000\,s, then the angular speed of the Earth is:

\omega = \frac{2\pi}{31536000\,s}

\omega = 1.992\times 10^{-7}\,\frac{rad}{s}

Now, we determine the linear speed:

v = (1.5\times 10^{8}\,km)\cdot \left(1.992\times 10^{-7}\,\frac{rad}{s} \right)

v = 29.88\,\frac{km}{s}

Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

6 0
2 years ago
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