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sammy [17]
3 years ago
12

What is the approximate number of galaxies in the known universe, assuming that the milky way has an average mass and that all o

f the mass in the known universe is in galaxies? the mass of the milky way is on the order of 1042 kg, and the mass of the known universe is about about 1053 kg?
Physics
1 answer:
igor_vitrenko [27]3 years ago
5 0
I think that the correct reporting of the given values are 10⁴² kg for the Milky Way and 10⁵³ kg for the whole universe. That would be logical in that case. Since it is mentioned that each galaxy has the same mass as the Milky Way, then the number of galaxies in the universe are:

Number of galaxies = 10⁵³ kg/10²⁴ kg = 10¹¹ or 100 trillion galaxies 
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Could You Take 2 Elements From The Periodic Table Of Elements, and possibly make one of other 116 elements that are on the table
LekaFEV [45]
That's what stars do all the time. 
For example, in the sun (and MOST other stars), deep down in the center
of the sun's core, two atoms of Hydrogen get squashed together so hard
that they blend into one atom of Helium AND release some energy. 
That's where the sun's energy all comes from.  It's called "nuclear fusion". 
It needs tremendous temperature and pressure to happen.  We know how
to do it, but we can't control it.  So far, the only thing we've ever been able
to use it for is Hydrogen bombs.   

There are 92 elements on the Periodic Table that are found in nature,
plus another 20 or so that have been made in the laboratory, but only
a few atoms of them.
5 0
3 years ago
Read 2 more answers
Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y
Gnesinka [82]
The representation of this problem is shown in Figure 1. So our goal is to find the vector \overrightarrow{R}. From the figure we know that:

\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}

From geometry, we know that:

\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}

Then using vector decomposition into components:

For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85

Therefore:

R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector \overrightarrow{R}, that is:
</span>
\left | \overrightarrow{R} \right |=&#10;\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span>\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}


6 0
3 years ago
A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga
BartSMP [9]

Answer:

Total work done in expansion will be 3.60\times 10^5J

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm =1.01\times 10^5Pa

So 2.10 atm =2.10\times 1.01\times 10^5=2.121\times 10^5Pa

Volume is increases from 3370 liter to 5.40 liter

So initial volume V_1=3.70liter

And final volume V_2=5.40liter

So change in volume dV=5.40-3.70=1.70liter

For isobaric process work done is equal to W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J

So total work done in expansion will be 3.60\times 10^5J

5 0
2 years ago
Find the dimension of density​
Irina-Kira [14]
Density = Mass divided by Volume
6 0
3 years ago
A 15 W electric shaver is used for 3 minutes. How much energy does it use in joules?​
borishaifa [10]
P=change in E/t
Change in E=p*t
=15*3
=45

The answer is 45J.
3 0
3 years ago
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