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Leto [7]
1 year ago
7

the choices below each describe the appearance of an h-r diagram for a different star cluster. which cluster is most likely to b

e located in the halo of our galaxy?
Physics
1 answer:
Kitty [74]1 year ago
7 0

The cluster that is most likely to be located in the halo of our galaxy is the diagram that shows main-sequence stars of every spectral type except O, along with a few giants and supergiants.

<h3>What are star clusters?</h3>

Star clusters are large collections of stars. Star clusters are classified into two types: Globular clusters are gravitationally bound groups of tens of thousands to millions of old stars.

Because of their location on the dusty spiral arms of spiral galaxies, they are sometimes referred to as galactic clusters. Stars in an open cluster share a common ancestor as they all formed from the same massive molecular cloud.

A typical spiral galaxy has a faint, extended stellar halo. A stellar halo is an essentially spherical population of stars and globular clusters thought to surround most disk galaxies and the cD class of elliptical galaxies. It should be noted that a halo is a spherical cloud of stars surrounding a galaxy. Astronomers have proposed that the Milky Way's halo is composed of two populations of stars.

Learn more about star on:

brainly.com/question/21379923

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2 years ago
A person shuffles across the rug and then place his finger near his friend's nose, causing a small spark that transfers about 10
sineoko [7]

Answer:

n = \frac{10^{-9}C}{1.6 x10^{-19}C}= 6.25 x10^{9} electrons \approx 6.3 x10^{9} electrons

Explanation:

For this case we know that the charge associated to the transferred electrons of the small spark is 10^{-19}C and we want to find the number of electrons transferred.

From the properties of matter we know that the charge of the electron is

q_e = |-1.6 x10^{-19}C |= 1.6 x10^{-19}C

And we know that the total charge is given by this expression:

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Where Q represent the charge, n the number of electrons and e the charge of the electron.

If we solve for n we got:

n = \frac{Q}{e}

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n = \frac{10^{-9}C}{1.6 x10^{-19}C}= 6.25 x10^{9} electrons \approx 6.3 x10^{9} electrons

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