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alekssr [168]
3 years ago
8

Light rays from stars bend toward smaller angles as they enter Earth's atmosphere. a. Explain why this happens using Snell's law

and the speed of light. b. Where are the actual stars in relation to their apparent position as viewed from the Earth's surface?

Physics
1 answer:
Grace [21]3 years ago
8 0

Answer:

Following are the answer to this question:

Explanation:

In option (a):

  • The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.  
  • Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.

In option (b):

  • Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.  
  • Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.

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Answer:

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Explanation:

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a radio station broadcast a frequency 90500 Hz. these radio waves travel at speed of 30000 m/s. what is the wavelength of the ra
Feliz [49]

Answer:

0.331m

Explanation:

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2 years ago
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two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel .calculate the equivalent series resist
Novosadov [1.4K]

Explanation:

Given that,

Two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel.

For series combination,

R_{eq}=R_1+R_2

For parallel combination,

\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}

When 6 ohm and 3 ohm are in series,

R_s=6+3\\\\R_s=9\ \Omega

When 6 ohm and 3 ohm are in paralle,

\dfrac{1}{R_p}=\dfrac{1}{6}+\dfrac{1}{3}\\\\R_p=2\ \Omega

So, the equivalent resistance in series combination is 9 ohms and in parallel combination it is 2 ohms.

6 0
2 years ago
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