Calculate the speed with which the moon orbits the earth given the distance from earth to moon as r = 3.84 · 108 m. (astronomers note that the true orbital period of the moon, is 27.3 earth days. interestingly, this would mean that there are approximately 13 months in a year. use the 27.3 days/month for t - the time required for one revolution in your calculation.)

a. 2.78x10^4m/s
b.2.41x10^4m/s
c.2.45x10^4m/s
d.1.02x10^3m/s

D. 1.02×10³ m/s

Explanation:

The radius of the moon's orbit is 3.84×10⁸ m.  So the circumference is:

C = 2πR

C = 2π (3.84×10⁸ m)

C = 2.41×10⁹ m

The period is 27.3 days.  Converting to seconds:

T = 27.3 days × (24 hrs / day)

T = 655.2 hrs × (3600 s / hr)

T = 2.36×10⁶ s

So the speed is:

v = C / T

v = (2.41×10⁹ m) / (2.36×10⁶ s)

v = 1.02×10³ m/s

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation:

temp rises, and expansion