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yan [13]
2 years ago
9

The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:

I = P 4πd2 . Suppose two stars of the same apparent brightness I are also believed to be the same size. The spectrum of one star peaks at 750 nm while the other peaks at 450 nm. Use Wien’s Law and the Stefan-Boltzmann equation to estimate their relative distances from u
Physics
1 answer:
kow [346]2 years ago
5 0

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!

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statuscvo [17]

Answer:

t = 0.33h = 1200s

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Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

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v1 = 55km/h

car 2:

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For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}

t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s

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6 0
2 years ago
Suppose an acorn with a mass of 3.17 g falls off a tree. At a particular moment during the fall, the acorn has a kinetic energy
8090 [49]

Answer:

potential energy decrease and kinetic energy increases.

Explanation:

the law of conservation of energy says that the total energy of a system is conserved, then:

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so, when the acorn fall the final potential energy is going to decrease because the potential energy is equal to:

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Image that the radiation emitted by the nitrogen at a frequency of 8.88×1014 Hz is absorbed by an electron in a molecule of meth
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Answer:

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Substituting,

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Therefore, the initial v1 = vcom (m1+m2) / m1 = 2.0 m/s x 6 = 12 m/s

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