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I am Lyosha [343]
3 years ago
9

While investigating Kirchhoff's Laws, you begin observing a blackbody, such as a star, from Earth using advanced technology that

can analyze spectra. While pointing it at the star with nothing between you and the star, you observe a full spectrum. You come back and repeat this same experiment a year later using the same star, except this time you observe an absorption spectrum. What is the most likely explanation for this
Physics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

the second time there is a gas between you and the star,

Explanation:

When you observe the star for the first time you do not have a given between you and the star, therefore you observe the emission spectrum of the same that is formed by lines of different intensity and position that indicate the type and percentage of the atoms that make up the star.

 When you observe the same phenomenon for the second time there is a gas between you and the star, this gas absorbs the wavelengths of the star that has the same energies and the atomisms and molecular gas, therefore these lines are not observed by seeing a series of dark bands,

The information obtained from the two spectra is the same, the type of atoms that make up the star

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During most of its lifetime, s star maintains an equilibrium size in which the inward force of gravity on each atom is balanced
frez [133]

Answer:

r_f=137493m

v=8638940m/s

Explanation:

During this process the mass M=2\times10^{30}Kg will be considered constant. We start from a radius r_i=7\times10^8m and a period T_i=30\ days=(30)(24)(60)(60)s=2592000s. The final period is T_f=0.1s.

Angular momentum <em>L</em> is conserved in this process. We can use the formula L=I\omega, where I is the momentum of inertia (which for a solid sphere is I=\frac{2mr^2}{5}) and \omega=\frac{2\pi }{T} is the angular velocity, so we can write the star's angular momentum as:

L=I\omega=\frac{2mr^2}{5}\frac{2\pi }{T}=\frac{4\pi mr^2 }{5T}

Since L_f=L_i we have:

\frac{4\pi mr_f^2 }{5T_f}=\frac{4\pi mr_i^2 }{5T_i}

Which can be simplified as:

\frac{r_f^2 }{T_f}=\frac{r_i^2 }{T_i}

Which means:

r_f=\sqrt{\frac{r_i^2 T_f}{T_i}}=r_i \sqrt{\frac{T_f}{T_i}}

Which for our values is:

r_f=r_i \sqrt{\frac{T_f}{T_i}}=(7\times10^8m) \sqrt{\frac{0.1s}{2592000s}}=137493m

And we calculate the speed of a point on the equator by dividing the final circumference over the final period:

v=\frac{C_f}{T_f}=\frac{2\pi r_f}{T_f}=\frac{2\pi (137493m)}{(0.1s)}=8638940m/s

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