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LekaFEV [45]
1 year ago
12

The vernal equinox is now just entering the constellation aquarius. In what constellation will it lie in the year a. D. 10000?.

Physics
1 answer:
Keith_Richards [23]1 year ago
8 0

The vernal equinox in the year AD 10000 will lie in the constellation Scorpio.

Once every 25800 years a full cycle of precession is completed. There are 12 constellations which means it takes 25800 / 12 = 2150 years to move from one constellation to the next one. The order in which the vernal equinox moves from Aquarius is Capricorn, Sagittarius, Scorpio, Libra etc.

Now the year is 2022 AD where the vernal equinox just entered constellation Aquarius. In 4172 AD it would have entered constellation Capricorn, then constellation  Sagittarius in 6322 AD, then constellation  Scorpio in 8472 AD and then constellation  Libra in 10622 AD. This means that in 10000 AD it is still in constellation  Scorpio and yet to enter constellation  Libra.

Therefore, In the year 10000 AD, the vernal equinox will lie in the constellation Scorpio.

To learn more about constellation Scorpio

brainly.com/question/4444209

#SPJ4

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Along one side of the square, the number of atoms that could fit =\frac {length\hspace{1mm}of\hspace{1mm}side}{diameter\hspace{1mm}of\hspace{1mm}atom}=\frac{9.5\times10^{-2}m}{4.95\times10^{-10}m}=1.91\times10^8 atoms

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consider the attached figure below:

The next layer of atoms are filled in the depressions of the first layer. The vertical distance between the two atoms would change. Consider an equilateral triangle drawn by joining the centers of the three atoms.

The atoms along the horizontal side would be aligned side by side; same as above =1.91\times10^8 atoms

The number of atoms along vertical side would vary. vertical distance between two atoms can be calculate using equilateral triangle as shown below.

an equilateral triangle has all the angles = 60^o

Let the vertical distance be y.

Then, y = 2r sin60^o

where r is the radius of each atom. 2r= 4.95\times10^{-10} m

y=4.95\times10^{-10} m\times sin60^o=4.28\times10^{-10} m

The number of atoms along the vertical side = \frac{9.5\times10^{-2}m}{4.28\times10^{-10}m}=2.21\times10^8 atoms

Total number of atoms in this kind of arrangement = 1.91\times10^8\times2.21\times10^8=4.23\times10^{16} atoms


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