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EleoNora [17]
1 year ago
9

The Sloan Digital Sky Survey includes many thousands of galaxies in its spectroscopic catalogs. How are the distances to most of

these galaxies estimated
Physics
1 answer:
Katarina [22]1 year ago
3 0

The distances to the most of these galaxies estimated from their red shifts and the application of Hubble's Law.

<h3>What is Hubble's Law?</h3>

The discovery in physical cosmology that galaxies are travelling away from Earth at rates proportionate to their distance is known as Hubble's law, often referred to as the Hubble-Lemaître law or Lemaître's law. In other words, they are travelling away from Earth more quickly the more away they are. The redshift of the galaxies—a shift in the light they produce toward the red end of the visible spectrum—has been used to calculate their velocities.

The Sloan Digital Sky Survey includes many thousands of galaxies in its spectroscopic catalogs. The distances to the most of these galaxies estimated

from their red shifts and the application of Hubble's Lawh

to learn more about Hubble's law go to - brainly.com/question/3050512

#SPJ4

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Two loudspeakers are 1.60 m apart. A person stands 3.00 m from one speaker and 3.50 m from the other. (a) What is the lowest fre
VMariaS [17]

Answer:

Explanation:

Given

Distance between two loud speakers d=1.6\ m

Distance of person from one speaker x_1=3\ m

Distance of person from second speaker x_2=3.5\ m

Path difference between the waves is given by

x_2-x_1=(2m+1)\cdot \frac{\lambda }{2}

for destructive interference m=0 I.e.

x_2-x_1=\frac{\lambda }{2}

3.5-3=\frac{\lambda }{2}

\lambda =0.5\times 2

\lambda =1\ m

frequency is given by

f=\frac{v}{\lambda }

where v=velocity\ of\ sound\ (v=343\ m/s)

f=\frac{343}{1}=343\ Hz

For next frequency which will cause destructive interference is

i.e. m=1 and m=2

3.5-3=\frac{2\cdot 1+1}{2}\cdot \lambda

\lambda =\frac{1}{3}\ m

frequency corresponding to this is

f_2=\frac{343}{\frac{1}{3}}=1029\ Hz

for m=2

3.5-3=\frac{5}{2}\cdot \lambda

\lambda =\frac{1}{5}\ m

Frequency corresponding to this wavelength

f_3=\frac{343}{\frac{1}{5}}

f_3=1715\ Hz                        

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3 years ago
1.
irga5000 [103]

Answer:

Explanation:

solution is found below

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