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3 years ago

14

Which of the following elements is commonly found in the Earth's crust, living matter, oceans, and atmosphere? A. carbon B. argo

n C. bromine D. zinc

2 answers:

jekas [21]3 years ago

8 0

The answer is carbon.

OverLord2011 [107]3 years ago

5 0

I'm pretty sure it's A. carbon

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givi [52]

Answer:

<u>For M84:</u>

M = 590.7 * 10³⁶ kg

<u>For M87:</u>

M = 2307.46 * 10³⁶ kg

Explanation:

1 parsec, pc = 3.08 * 10¹⁶ m

The equation of the orbit speed can be used to calculate the doppler velocity:

$v = \sqrt{\frac{GM}{r} }$

making m the subject of the formula in the equation above to calculate the mass of the black hole:

$M = \frac{v^{2} r}{G}$ .............(1)

<u>For M84:</u>

r = 8 pc = 8 * 3.08 * 10¹⁶

r = 24.64 * 10¹⁶ m

v = 400 km/s = 4 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

$M = \frac{( 4*10^{5}) ^{2} *24.64* 10^{16} }{6.674 * 10^{-11} }$

M = 590.7 * 10³⁶ kg

<u>For M87:</u>

r = 20 pc = 20 * 3.08 * 10¹⁶

r = 61.6* 10¹⁶ m

v = 500 km/s = 5 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

$M = \frac{( 5*10^{5}) ^{2} *61.6* 10^{16} }{6.674 * 10^{-11} }$

M = 2307.46 * 10³⁶ kg

The mass of the black hole in the galaxies is measured using the doppler shift.

The assumption made is that the intrinsic velocity dispersion is needed to match the line widths that are observed.

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3 years ago

An electron with charge −e and mass m moves in a circular orbit of radius r around a nucleus of charge Ze, where Z is the atomic

shepuryov [24]

Answer:

$v=\sqrt{\frac{kZe^2}{mr}}$

Explanation:

The electrostatic attraction between the nucleus and the electron is given by:

$F=k\frac{(e)(Ze)}{r^2}=k\frac{Ze^2}{r^2}$ (1)

where

k is the Coulomb's constant

Ze is the charge of the nucleus

e is the charge of the electron

r is the distance between the electron and the nucleus

This electrostatic attraction provides the centripetal force that keeps the electron in circular motion, which is given by:

$F=m\frac{v^2}{r}$ (2)

where

m is the mass of the electron

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Combining the two equations (1) and (2), we find

$k\frac{Ze^2}{r^2}=m\frac{v^2}{r}$

And solving for v, we find an expression for the speed of the electron:

$v=\sqrt{\frac{kZe^2}{mr}}$

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2 years ago