Answer:
1. Main Sequence - middle life 17
2. red
3. blue
4. White dwarf stars are much hotter than Red Supergiants 15. List the color of the stars from hottest to coldest: Blue, White, Yellow, Orange, Red 16.
5. red giants
Explanation:
Main sequence stars have a Morgan-Keenan luminosity class labeled V. red giant and supergiant stars (luminosity classes I through III) occupy the region above the main sequence. They have low surface temperatures and high luminosities which, according to the Stefan-Boltzmann law, means they also have large radii. White dwarf stars are much hotter than Red Supergiants 15. List the color of the stars from hottest to coldest: Blue, White, Yellow, Orange, Red 16. The hottest stars are the blue stars. A star appears blue once its surface temperature gets above 10,000 Kelvin, or so, a star will appear blue to our eyes. The lowest temperature stars are red while the hottest stars are blue. Astronomers are able to measure the temperatures of the surfaces of stars by comparing their spectra to the spectrum of a black body. Most stars, including the sun, are "main sequence stars," fueled by nuclear fusion converting hydrogen into helium. ...
As stars begin to die, they become giants and supergiants (above the main sequence).
Answer:
El flujo del campo U a través de dicha superficie es ![\Phi_U=4\pi r^2K](https://tex.z-dn.net/?f=%5CPhi_U%3D4%5Cpi%20r%5E2K)
Explanation:
El flujo vectorial de un campo U a través de una superficie S está dado por
![\Phi_U=\int_S \vec{U} \cdot \hat{n} \ dS](https://tex.z-dn.net/?f=%5CPhi_U%3D%5Cint_S%20%5Cvec%7BU%7D%20%5Ccdot%20%5Chat%7Bn%7D%20%5C%20dS)
donde
representa el vector unitario normal a la superficie. Teniendo en cuenta que el campo se encuentra en dirección radial, se tiene
![\Phi_U=\int_S K\hat{r} \cdot \hat{r} \ dS = K\int_S dS = KS=K*4\pi r^2](https://tex.z-dn.net/?f=%5CPhi_U%3D%5Cint_S%20K%5Chat%7Br%7D%20%5Ccdot%20%5Chat%7Br%7D%20%5C%20dS%20%3D%20K%5Cint_S%20dS%20%3D%20KS%3DK%2A4%5Cpi%20r%5E2)
The answer is A. Atoms. You're welcome. ;)
About 90 percent of the stars in space are Main Sequence stars.<em> (c)</em>
Answer:4.39 s
Explanation:
Given
initial velocity ![u=0](https://tex.z-dn.net/?f=u%3D0)
acceleration ![a=12.8 m/s^2](https://tex.z-dn.net/?f=a%3D12.8%20m%2Fs%5E2)
velocity acquired by sled in
time
![v=0+at](https://tex.z-dn.net/?f=v%3D0%2Bat)
![v=12.8t_1](https://tex.z-dn.net/?f=v%3D12.8t_1)
distance traveled by sled in ![t_1 s](https://tex.z-dn.net/?f=t_1%20s)
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
![(12.8t_1)^2-0=2\times 12.8\times s_1](https://tex.z-dn.net/?f=%2812.8t_1%29%5E2-0%3D2%5Ctimes%2012.8%5Ctimes%20s_1)
![s_1=6.4\cdot t_1^2](https://tex.z-dn.net/?f=s_1%3D6.4%5Ccdot%20t_1%5E2)
distance traveled in
time with velocity ![v=12.8t_1](https://tex.z-dn.net/?f=v%3D12.8t_1)
![s_2=v\times t_2](https://tex.z-dn.net/?f=s_2%3Dv%5Ctimes%20t_2)
![s_2=12.8\times t_1\times t_2](https://tex.z-dn.net/?f=s_2%3D12.8%5Ctimes%20t_1%5Ctimes%20t_2)
![s_2=12.8\cdot t_1\cdot t_2](https://tex.z-dn.net/?f=s_2%3D12.8%5Ccdot%20t_1%5Ccdot%20t_2)
![s_1+s_2=5.37\times 10^3](https://tex.z-dn.net/?f=s_1%2Bs_2%3D5.37%5Ctimes%2010%5E3)
----1
![t_1+t_2=97.7 s](https://tex.z-dn.net/?f=t_1%2Bt_2%3D97.7%20s)
![t_2=97.7-t_1](https://tex.z-dn.net/?f=t_2%3D97.7-t_1)
substitute the value of
in 1
we get
![6.4t_1^2-1250.56t_1+5370=0](https://tex.z-dn.net/?f=6.4t_1%5E2-1250.56t_1%2B5370%3D0)
thus ![t_1=\frac{1250.56-1194.33}{12.8}=4.39 s](https://tex.z-dn.net/?f=t_1%3D%5Cfrac%7B1250.56-1194.33%7D%7B12.8%7D%3D4.39%20s)
![t_1=4.39 s](https://tex.z-dn.net/?f=t_1%3D4.39%20s)