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schepotkina [342]

3 years ago

12

Is the star with the longest total lifetime Also the farthest from earth? Explain.

1 answer:

Stels [109]3 years ago

6 0

E) No. Ollie will shine for 30 Billion years but is only 10,000 LY from Earth.
F) No. Cosmo will shine for 3 Million years but is 10 Billion LY from Earth.
G) Yes. Ollie is only 10.000 LY away but will shine for 30 Billion years.
Ga) No. Stars such as Cosmo shine for 3 Million years.
Gb) If Cosmo was also 3 Million LY away we would see it now.

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As the average kinetic energy of a substance increases, the tempature of the sample

taurus [48]

I have no clue. Sorry bro

6 0

3 years ago

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In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a

IRISSAK [1]

A. $4.64\cdot 10^{11}m$

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

$v=\frac{2\pi r}{T}$

where we have:

$v=30,000 km/s = 3\cdot 10^7 m/s$ is the orbital speed

r is the orbital radius

$T=27 h \cdot 3600 =97,200 s$ is the orbital period

Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

$r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m$

B. $6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s$

The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

$m\frac{v^2}{r}=\frac{GMm}{r^2}$

where

m is the mass of the clumps of matter

G is the gravitational constant

M is the mass of the black hole

Solving the formula for M, we find the mass of the black hole:

$M=\frac{v^2 r}{G}=\frac{(3\cdot 10^7 m/s)^2(4.64\cdot 10^{11} m)}{6.67\cdot 10^{-11}}=6.26\cdot 10^{36}kg$

and considering the value of the solar mass

$M_s = 2\cdot 10^{30}kg$

the mass of the black hole as a multiple of our sun's mass is

$M=\frac{6.26\cdot 10^{36} kg}{2\cdot 10^{30} kg}=3.13\cdot 10^6 M_s$

C. $9.28\cdot 10^9 m$

The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by

$R=\frac{2MG}{c^2}$

where M is the mass of the black hole and c is the speed of light.

Substituting numbers into the formula, we find

$R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m$

8 0

3 years ago

To measure one wavelength, you would measure the distance between?

makvit [3.9K]

A and B, one wavelength is crest too crest

8 0

3 years ago

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Scrat [10]

Answer:

c

Explanation:

4 0

3 years ago

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A white blood cell has a diameter of approximately 12 micrometers, or 0.012 um. A model represents its diameter as 24 um. What i

GuDViN [60]

Answer:

2:1

Explanation:

It is given that,

The diameter of a white blood cell is 12 micrometers

The diameter of white blood cell according to a model is 24 micrometers.

We need to find the ratio of model size to actual size. It can be done as follows :

$R=\dfrac{\text{model size}}{\text{actual size}}\\\\R=\dfrac{24\ \mu m}{12\ \mu m}\\\\R=\dfrac{2}{1}$

Hence, the ratio of model size to actual size is 2:1.

8 0

3 years ago