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.)

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%

the answer is a 3.8cm

explanation: