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Misha Larkins [42]
3 years ago
6

This theory lost its appeal when astronomers discovered quasars and cosmic background radiation.

Physics
2 answers:
Vesnalui [34]3 years ago
4 0

Answer:

The steady state theory

Explanation:

According to the steady state model of the universe, the universe is continuously expanding while maintaining average density as constant. According to the theory, matter is continuously being formed and creating new stars and galaxies while the old ones are becoming un-observable. There is no beginning or end of time.

On the other hand, the big bang theory describes that the entire matter of the universe was confined in a point and with the big bang, this point started expanding. This happened about 13.8 billion years ago. Of course the density and temperature of the universe was very high at the beginning. In the initial few minutes of the big bang, the universe expanded rapidly which resulted in the decrease in its density and temperature. After million years, the universe transparent for radiation to freely travel through space. Cosmic Microwave background is the remnant of this early universe and is a strong evidence of the Big Bang Theory. The number of quasars were greater in past than present which shows that universe is evolving with time.

JulijaS [17]3 years ago
3 0
<span>the Steady State Theory is the answer to this I believe!</span>
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When we do dimensional analysis, we do something analogous to stoichiometry, but with multiplying instead of adding. Consider th
Ksivusya [100]

Answer:

The  dimension is  D =  L ^{2} T^{-1}

Explanation:

From the question we are told that

     J  =  -D \frac{dn}{dx}

Here  [J] = \frac{1}{L^2 T}

       [n] =\frac{1}{L^3}

        [x] = L

So

    \frac{1}{L^2 T} =  -D \frac{d(\frac{1}{L^3})}{d[L]}

Given that the dimension represent the unites of  n and  x then the differential  will not effect on them

So

\frac{1}{L^2 T} =  -D \frac{(\frac{1}{L^3})}{[L]}

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=>   D =  L ^{2} T^{-1}

   

5 0
3 years ago
a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t
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Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

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The pressure at the depth of 40 m: P1=1atm+dρg

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ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

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