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6 March, 00:24

Because of interstellar dust, astronomers can see at most about 5 kpc into the disk of the galaxy at visual wavelengths. What percentage of the galactic disk does that include? (Hint: Consider the area of the entire disk versus the area visible from Earth. The diameter of the Milky Way Galaxy is approximately 25 kpc.)

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  1. 6 March, 01:36
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    96%

    Explanation

    Let A the total area of the galaxy, is modeled as a disc:

    A = πR^2 = π (25 kpc) ^2

    And let a be the area that astronomers are able to see:

    a = πr^2 = π (5 kpc) ^2

    The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:

    P = 100 a/A = (5/25) ^2 = 100/25 = 4%

    Therefore, the percentage of the galaxy not included, i. e. not seen is:

    (100-4) % = 96%
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