Respuesta :
Answer:
The supernova is a fraction of 0.7 of the size of the observable universe which is 10 billion light years away.
Explanation:
Apparent brightness(AB) = luminosity(L) / 4πr²
∴ Luminosity(L) = Apparent brightness(AB) × 4πr²
Ratio of the luminosity of cepheid and supernova with the same brightness will be written as;
[tex]\frac{L_{cepheid}}{L_{supernova}} = \frac{AB_{cepheid}(4\pi r^2)}{AB_{supernova}(4\pi r^2)}[/tex]
[tex]= \frac{AB_{cepheid}(r^2_{cepheid})}{AB_{supernova}(r^2_{supernova})}[/tex]
[tex]\frac{(r^2_{cepheid})}{(r^2_{supernova})}[/tex]
Therefore, [tex]r_{supernova} = r_{cepheid}\sqrt{\frac{luminosity_{cepheid}}{luminosity_{supernova}} }[/tex] ...................(a)
Since we already have [tex]r_{cepheid}[/tex] as 100 million light year, substitute for luminosity of supernova and luminosity of cepheid as 10⁸[tex]L_{sun}[/tex] and 10,000[tex]L_{sun}[/tex] respectively in equation (a)
[tex]r_{supernova} = 100 million \sqrt{\frac{10^8L_{sun}}{10,000L_{sun}} }[/tex] ==> 10 billion light year
the ratio of the distance to the size of the universe:
Supernova / Observable Universe = (10 / 14) billion light years
Supernova / Observable Universe = 0.7
The supernova is a fraction of the size of 0.7 of the size of the viewable universe that is 10 bn light years away.
What is supernova and how far is it. ?
The supernova is a very powerful and luminous Steller explosion and is a transitional astronomical event and take place ducting the evolution and is one of the stages of the massive star. Triggered by the leakage of the gases.
The distance of the fading of a white dwarf supernova of brightness is 108 sun and has brightness as comparable to that star of the brightest Cepheid. The star has a variable of 100 million light years form earth. Thus distance of the nearest supernova is 50 billion light years and the observable is 10 billion.
Find out more information about the fading white.
brainly.com/question/15726471.
