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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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4 years ago
Two obiect accumulated a charge of
tamaranim1 [39]

Answer:

A. 181.24 N

Explanation:

The magnitude of hte electrostatic force between two charged objects is given by the equation

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

In this problem, we have:

q_1=4.5\mu C=4.5\cdot 10^{-6}C is the magnitude of the 1st charge

q_2=2.8\mu C=2.8\cdot 10^{-6}C is the magnitude of the 2nd charge

r = 2.5 cm = 0.025 m is the separation between the charges

Therefore, the magnitude of the electric force is:

F=\frac{(9\cdot 10^9)(4.5\cdot 10^{-6})(2.8\cdot 10^{-6})}{(0.025)^2}=181.44 N

So, the closest answer is

A) 181.24 N

3 0
4 years ago
A police car parked on the side of the highway emits a 1200 Hz sound that bounces off a vehicle farther down the highway and ret
emmasim [6.3K]

f' = frequency observed by the police car after sound reflected from the vehicle and comes back to police car = 1250 Hz

f = frequency emitted by the police car  = 1200 Hz

V = speed of sound = 340 m/s

v = speed of vehicle = ?

frequency observed by the police car is given as

f' = f (V + v)/(V - v)

inserting the values in the above equation

1250 = 1200 (340 + v)/(340 - v)

v = 6.9 m/s

4 0
4 years ago
Read 2 more answers
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