what is the hubble classification for a spiral galaxy with a small nuclear bulge and wide loosely wound spiral arms?

2.17 x 10^8 m/s

Explanation:

Angle of incidence, i = 30 degree, refractive index of mineral oil, n = 1.38

Let r be the angle of refraction.

By use of Snell's law

n = Sin i / Sin r

Sin r = Sin i / n

Sin r = Sin 30 / 1.38

Sin r = 0.3623

r = 21.25 degree

Let the speed of light in oil be v.

By the definition of refractive index

n = c / v

Where c be the speed of light

v = c / n

v = ( 3 x 10^8) / 1.38

v = 2.17 x 10^8 m/s