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inna [77]
3 years ago
12

The following figures give the approximate distances of five galaxies from earth. rank the galaxies based on the speed with whic

h each should be moving away from earth due to the expansion of the universe, from fastest to slowest.
Physics
1 answer:
Tju [1.3M]3 years ago
5 0
The galaxies are so far from the Earth, and their spectra so extremely
red-shifted, that I'm not able to see any of the items on the list.

Estimates of the Hubble constant still cover a wide range. 
Let's assume that it's  70 km/sec per megaparsec, or
about 21.5 km/sec per million light years.

With that factoid, the speed of recession of each galaxy on your
invisible list is roughly

       (21.5 km/sec) x (distance to the galaxy) / (1 million light years) .

You'll find ... if it's important enough to you for you to carry out the work ...
that the farthest galaxy is the fastest, the nearest one is the slowest,
and the others fall similarly in line.

In other words:

No matter where we look in the universe, and no matter
in what direction we look, we observe that:

-- all distant galaxies are moving away from us
and
-- the farther a galaxy already is from us, the faster
it's moving away from us.

This observation could have been enough to give us
a giant inferiority complex, or to cause us to go brush
our teeth and rub on some deodorant.
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You push a 55.0 kg block to the left with a 245 N force along a frictionless, level surface. The block starts
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2 years ago
What is the correct formula for finding the frequency of an electromagnetic wave
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Answer:
f = c / λ

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3 0
3 years ago
A voltaic cell is constructed with two Zn2+-Zn electrodes, where the half-reaction is Zn2+ + 2e− → Zn (s) E° = -0.763 V The conc
wariber [46]

Answer : The cell emf for this cell is 0.077 V

Solution :

The balanced cell reaction will be,  

Oxidation half reaction (anode):  Zn(s)\rightarrow Zn^{2+}+2e^-

Reduction half reaction (cathode):  Zn^{2+}+2e^-\rightarrow Zn(s)

In this case, the cathode and anode both are same. So, E^o_{cell} is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}{diluted}}{[Zn^{2+}{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Zn^{2+}{diluted}] = 0.0111 M

[Zn^{2+}{concentrated}] = 4.50 M

Now put all the given values in the above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{0.0111M}{4.50M}

E_{cell}=0.077V

Therefore, the cell emf for this cell is 0.077 V

7 0
3 years ago
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