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Sagot :
To determine which star will stay on the main sequence for the longest time, we need to understand the relationship between the mass of a star and the duration of its stay on the main sequence. The time a star remains on the main sequence is inversely proportional to the cube of its mass. This means that stars with lower masses will have longer lifespans on the main sequence compared to more massive stars.
Given the initial masses of the stars in solar masses:
- Star 1: 9 solar masses
- Star 2: 0.9 solar masses
- Star 3: 5 solar masses
- Star 4: 0.3 solar masses
Since the duration on the main sequence is inversely proportional to the cube of the mass, we need to find which star has the smallest mass, because that star will have the longest time on the main sequence.
Let's review the given masses:
- Star 1 has a mass of 9 solar masses.
- Star 2 has a mass of 0.9 solar masses.
- Star 3 has a mass of 5 solar masses.
- Star 4 has a mass of 0.3 solar masses.
Among these masses, 0.3 solar masses is the smallest. Therefore, the star with this mass will stay on the main sequence for the longest time.
Thus, Star 4 will stay on the main sequence the longest.
[tex]\[ \boxed{\text{Star 4}} \][/tex]
Given the initial masses of the stars in solar masses:
- Star 1: 9 solar masses
- Star 2: 0.9 solar masses
- Star 3: 5 solar masses
- Star 4: 0.3 solar masses
Since the duration on the main sequence is inversely proportional to the cube of the mass, we need to find which star has the smallest mass, because that star will have the longest time on the main sequence.
Let's review the given masses:
- Star 1 has a mass of 9 solar masses.
- Star 2 has a mass of 0.9 solar masses.
- Star 3 has a mass of 5 solar masses.
- Star 4 has a mass of 0.3 solar masses.
Among these masses, 0.3 solar masses is the smallest. Therefore, the star with this mass will stay on the main sequence for the longest time.
Thus, Star 4 will stay on the main sequence the longest.
[tex]\[ \boxed{\text{Star 4}} \][/tex]
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