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ivanzaharov [21]
2 years ago
11

Most stars are main-sequence stars, a group of stars for which size, mass, surface temperature, and radiated power are closely r

elated. The sun, for instance, is a yellow main sequence star with a surface temperature of 5800 K. For a main-sequence star whose mass M is more than twice that of the sun, the total radiated power, relative to the sun, is approximately P/Psun = 1.5(M/Msun)3.5. The star Regulus A is a bluish main-sequence star with mass 3.8Msun and radius 3.1Rsun. What is the surface temperature of Regulus A?
Physics
1 answer:
quester [9]2 years ago
4 0

Answer:

the surface temperature of Regulus A is 11724.13 K

Explanation:

Given the data in the question;

Sun's surface temperature T = 5800 K

total radiated power, relative to the sun is; P/P_{sun = ( 1.5(M/M_{sun )^{3.5

The star Regulus A is a bluish main-sequence star with mass 3.8M_{sun  and radius 3.1R_{sun .

First, we determine the value power emitted by the sun or sun as follows;

P = eσAT⁴

where P is the power, e is surface emissivity, σ is Stefan Boltzmann, A is area and T is temperature.

so, lets assume emissivity of star and sun is same;

let p be power related to star and p_{sun be power related sun.

Ratio of power radiated by star and power radiated by sun;

P/P_{sun = eσAT⁴ / eσA_{sunT_{sun⁴

we know that AREA A = πR²

we input the formula for area

P/P_{sun = eσ(πR²)T⁴ / eσ(π(R_{sun)²)T_{sun⁴

such that we now have;

P/P_{sun = R²T⁴ / R_{sun²T_{sun⁴

given that P/P_{sun = ( 1.5(M/M_{sun )^{3.5, we substitute

( 1.5(M/M_{sun )^{3.5 = R²T⁴ / R_{sun²T_{sun⁴

we find temperature of the star T

T = 5800  × [ 1.5(M/M_{sun)^{3.5 (R_{sun²/R²)]^{1/4

Given that;  mass M is 3.8M_{sun and radius R is 3.1R_{sun .

we substitute

T = 5800  × [ 1.5(3.8M_{sun/M_{sun)^{3.5 (R_{sun²/( 3.1R_{sun )²)]^{1/4

T = 5800  × [ 1.5(3.8)^{3.5 ( 1/( 3.1)²)]^{1/4

T = 5800  × [ 1.5( 106.9652 ) ( 1/(9.61) ]^{1/4

T = 5800  × [ 16.69592 ]^{1/4

T = 5800  ×  2.02140152

T = 11724.13 K

Therefore, the surface temperature of Regulus A is 11724.13 K

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5. Principal Rodriguez has asked all students exit the building by 2:45 pm! If you want to exit
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With the Pythagoras Theorem we can find that the distance traveled is:

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The distance is the length of the path between two points and it is a scalar magnitude so if the path changes direction the Pythagorean theorem should be used

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Where d is the distance, (x,y,z) is the interes point, (x₀,y₀,z₀) is de reference point.

In this case, let's set a reference system in the lower part of the school, take the z-axis as vertical and set the point of arrival at as the reference (0, 0, 0).

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In conclusion using the Pythagoras Theorem we can find that the distance traveled is 39 m

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brainly.com/question/7942332

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