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ValentinkaMS [17]
3 years ago
11

How is the Milky Way changing? A. Large numbers of stars are being born inside nebulas. B. The Milky Way does not change. C. Lar

ge numbers of stars are dying inside nebulas. D. Large numbers of stars are undergoing blue-shift.
HELP ME PLEASE!
Physics
1 answer:
Over [174]3 years ago
7 0

Answer:

D. Large numbers of stars are undergoing blue-shift.

Explanation:

The Milky Way is changing because a large numbers of stars are undergoing blue-shift.

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A mass m = 14 kg is pulled along a horizontal floor with NO friction for a distance d =5.7 m. Then the mass is pulled up an incl
frosja888 [35]

Answer:

W ≅ 292.97 J

Explanation:

1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)

Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;

W = (Fcosθ)d

Given that:

Tension of the force = 62 N

angle of incline θ =  34°

distance d =5.7 m.

Then;

W = 62 × cos(34) × 5.7

W = 353.4 cos(34)

W = 353.4 × 0.8290

W = 292.9686 J

W ≅ 292.97 J

Hence,  the work done by tension before the block goes up the incline = 292.97 J

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3 years ago
Calculate the gravitational potential energy of a 1200 kg car at the top of a hill that is 42 m high.
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It IS <span>PE = (1200 kg)(9.8 m/s²)(42 m) = 493,920 J </span>
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. As we increase the quantum number of an electron in a one-dimensional, infinite potential well, what happens to the number of
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Answer:

It increases.

Explanation:

For the electron to escape the photon needs energy is equal to the difference between initial and its non quantised region energy , then only it will be able to escape finite well.

E ∝ n^2

n= energy state quantum number

so if , n increases maximum point of probability density increases.

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An object moves in uniform circular motion what is true regarding the force on the object
PtichkaEL [24]
The force on the object has a constant strength, but its direction
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4 0
3 years ago
Read 2 more answers
The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:
kow [346]

Answer:

\frac{d_{1}}{d_{2}}=0.36

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]} (1)

So, the temperature of the first and the second star will be:

T_{1}=3866.7 K

T_{2}=6444.4 K

Now the relation between the absolute luminosity and apparent brightness  is given:

L=l\cdot 4\pi r^{2} (2)

Where:

  • L is the absolute luminosity
  • l is the apparent brightness
  • r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}

So the relative distance between both stars will be:

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}} (3)

The Boltzmann Law says, L=A\sigma T^{4} (4)

  • σ is the Boltzmann constant
  • A is the area
  • T is the temperature
  • L is the absolute luminosity

Let's put (4) in (3) for each star.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}

As we know both stars have the same size we can canceled out the areas.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=0.36

I hope it helps!

5 0
3 years ago
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