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alekssr [168]
2 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]2 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

Given,

Periscope uses 45-45-90 prisms with total internal reflection adjacent to 45°.

refractive index of water, n_a = 1.33

refractive index of glass, n_g = 1.52

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A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol
qaws [65]

Answer:

Part a)

v = \sqrt{v_o^2 + g^2t^2}

Part b)

d = \frac{1}{2}gt^2

Part c)

v_f = v_o - gt

Part d)

d = \frac{1}{2}gt^2

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

v_y = -gt

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{v_o^2 + g^2t^2}

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

d = \frac{1}{2}gt^2

Part c)

If helicopter is rising upwards with uniform speed

then final speed of the package after time t is given as

v_f = v_i + at

v_f = v_o - gt

Part d)

distance from helicopter

d = \frac{1}{2}gt^2

8 0
2 years ago
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The angle of refraction would be further less 
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Answer:

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netineya [11]
Refer to the diagram shown below.

v = the tangential speed.
r = the radius of the horizontal circle.
T = tension in the string.
θ =  the angle that the string makes with the vertical
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F =  centripetal force
l = the length of the string

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r = l sin θ

The centripetal acceleration is
a= \frac{v^{2}}{r} = \frac{v^{2}}{l \,sin \theta}
The centripetal force is
F = \frac{m v^{2}}{l \, sin \theta}

For vertical force balance,
T cosθ = mg                              (1)
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