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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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Contact [7]

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

d=\sqrt{(x-x_o)^2+(y-y_o)^2}   (1)

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6 0
3 years ago
A fireman standing on a 15 m high ladder
ludmilkaskok [199]

Answer:

31.6 m/s

Explanation:

Mass is conserved, so the mass flow at the outlet of the pump equals the mass flow at the nozzle.

m₁ = m₂

ρQ₁ = ρQ₂

Q₁ = Q₂

v₁A₁ = v₂A₂

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Now use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since h₁ = 0 and P₂ = 0:

P₁ + ½ ρ v₁² = ½ ρ v₂² + ρgh₂

Writing v₁ in terms of v₂:

P₁ + ½ ρ (v₂ d₂²/d₁²)² = ½ ρ v₂² + ρgh₂

P₁ + ½ ρ (d₂/d₁)⁴ v₂² = ½ ρ v₂² + ρgh₂

P₁ − ρgh₂ = ½ ρ (1 − (d₂/d₁)⁴) v₂²

Plugging in values:

579,160 Pa − (1000 kg/m³)(9.8 m/s²)(15 m) = ½ (1000 kg/m³) (1 − (1.99 in / 3.28 in)⁴) v₂²

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8 0
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To answer this question, it helps enormously if you know
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Looking at the formula, you can see that momentum is directly
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If the car's momentum is 20,000 kg-m/s now, then after its speed
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6 0
3 years ago
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