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BlackZzzverrR [31]
3 years ago
15

Estimate the diameter of the supergiant star Betelgeuse, using its angular diameter of about 0.05 arcsecond and distance of abou

t 600 light years.
Express your answer to three significant figures and include the appropriate units.
Mathematics
2 answers:
8090 [49]3 years ago
7 0

<span>0.05 arc-second = 1 degree/72000 = (pi radians)/(180*72000) = 2.424 x 10^(-7) radians</span>

<span>The distance is roughly: </span>

<span>R*(theta) = (600 light-years)*2.424 x 10^(-7) = 0.00014544 light-years = 1.275 light-hours = (3600 seconds)*(3 x 10^8 m/s)*(1.275) = 1.38 x 10^12 meters.</span>

forsale [732]3 years ago
5 0

Answer:

Diameter of the star, d=1.38\times 10^{12}\ m

Step-by-step explanation:

Given that,

Angular diameter of the star, angle, \theta=0.05\ arcsecond=2.42\times 10^{-7}\ radian

Distance, D=600\ ly=5.67\times 10^{18}\ m

We need to find the diameter of the supergiant star Betelgeuse. The relationship between the diameter of star and angle subtended is given by :

\theta=\dfrac{d}{D}, d = diameter

d=\theta\times D

d=2.42\times 10^{-7}\times 5.67\times 10^{18}

d=1.37214\times 10^{12}\ m

or

d=1.38\times 10^{12}\ m

Hence, this is the required solution.

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Step-by-step explanation:

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                                         = \frac{1}{2}(\text{GE+OF)}\times (\text{OG})

                                         = \frac{1}{2}(3+6)\times 4

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Area of trapezoid GCDE = \frac{1}{2}(\text{GC+DE)}\times (\text{GE})

                                         = \frac{1}{2}(7+2)\times 3

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Area of triangle AFH = \frac{1}{2}(\text{Base})\times (\text{Height})

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