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Arlecino [84]
3 years ago
14

When scientists observed the light from stars and galaxies, they noticed that their color shifted toward the end of the visible

color spectrum.
A. violet
B. Green
C. Red
D. Blue
Physics
2 answers:
kakasveta [241]3 years ago
8 0
<h2>Right answer: Red  </h2>

When we talk about the visible electromagnetic spectrum, we know it starts in <u>violet-blue</u> and ends in <u>red</u>.

Now, in this context the astronomer Edwin Powell Hubble observed several celestial bodies, and when obtaining <u>the spectra of distant galaxies</u> he observed that the spectral lines were displaced towards the red one (red shift), whereas <u>the nearby stars</u> showed a spectrum displaced to the blue one.

From there, Hubble deduced that the farther the galaxy is, the more redshifted it is in its spectrum, and noted that all galaxies are "moving away from each other with a speed that increases with distance".  

In other words: In the past the distance between two galaxies was smaller than at present, being this <u>the proof that the universe is expanding </u>(like a balloon expands when it is filled with air).

It should be noted that the redshift is not produced by the relative movement of the galaxies with each other.

<h2>This effect is due to the own expansion of the space among the galaxies. </h2>
lakkis [162]3 years ago
7 0
Color shifted towards the end is C red
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When did robert fulton invent the steamboat
bekas [8.4K]

Answer:

1807

Explanation:

Robert Fulton (1765–1815) was an American engineer and inventor who is widely known for developing a commercially successful steamboat called Clermont. In 1807, that steamboat took passengers from New York City to Albany and back again, a round trip of 300 miles, in 62 hours.

7 0
3 years ago
Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy
8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV

Energy of the second orbit in H atom .

E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV

Energy of the third orbit in H atom .

E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV

Energy of the fifth orbit in H atom .

E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV

Energy of the sixth orbit in H atom .

E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV

Energy of the seventh orbit in H atom .

E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV

During an absorption of energy electron jumps from lower state to higher state.So,  absorption will take place in :

1) n = 2 → n = 3

2) n=  5 → n = 6

Energy absorbed when: n = 2 → n = 3

E=E_3-E_2

E=(-1.51 eV) -(-3.40 eV)=1.89 eV

Energy absorbed when: n = 5 → n = 6

E'=E_6-E_5

E'=(-0.378 eV)-(-0.544 eV) =0.166 eV

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0
3 years ago
Catalytic ozone destruction occurs in the stratosphere by a two-step reaction:
Sladkaya [172]
Catalytic ozone destruction occurs in the stratosphere where the reactions involving bromine, chlorine, hydrogen, nitrogen and oxygen gases form compounds that destroy the ozone layer. The reactions uses a catalyst (speeds up the reaction) in a two step reaction. considering chlorine the reactions appears as follows;
step 1
 Cl + O3 = ClO + O2 
 step 2
ClO + O = Cl + O2 
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7 0
3 years ago
A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 29.6 m/s. At the same time, it h
inessss [21]

If the vertical component is 29.6 m/s down, and the horizontal component
is 54.8 m/s parallel to the surface, then the magnitude of the slanty vector is

   √(29.6² + 54.8²) = √(876.16 + 3003.04) = √3879.2  =  62.28 m/s .

That's 139 mph !  Wow !

6 0
3 years ago
What is the range of a ball thrown horizontally at 12 m/s if its time of flight is 3.0 s?
vladimir1956 [14]

Answer:

6

Explanation:

because I did this assignment, :) your welcome

Next time do it by yourself, but here's the answer kid

7 0
3 years ago
Read 2 more answers
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