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nataly862011 [7]
3 years ago
5

Laboratory experiments here on Earth have determined that each element in the periodic table emits photons only at certain wavel

engths. These photons are apparent to astronomers as either emission or absorption lines in the spectrum of an astronomical object, like stars in galaxies. Astronomers can, by measuring the position of these spectral lines, determine which elements are present in the star itself or along the line of sight. When astronomers perform this analysis, they note that for most astronomical bodies, the observed spectral lines are all shifted to longer (redder) wavelengths. This is known as cosmological redshift and is analogous to the Doppler redshift. What does this redshift indicate to us about the universe in general?
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0
The only thing we know about so far that can shift wavelengths of light
to longer wavelengths is when the source of the light is moving away
from the observer.

When we look at the light from distant galaxies, the light from them is
always shifted to longer wavelengths than it SHOULD have. 

AND ... The farther away from us a galaxy IS, the MORE its light is
shifted to wavelengths longer than it should have.

So far, this indicates to us that the whole universe is expanding.
That's the only way to understand what we see, because that's
the only thing we know of that can shift light to longer wavelengths.


By the way ... The most interesting thing about these observations
and measurements is:  When astronomers see this light from distant
galaxies and measure the wavelengths, how do they know how far
the wavelengths shifted ?  How do they know what the wavelengths
SHOULD be ?
 
I'll leave you to read about that in the next few years.
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A wooden block with mass 1.60 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 30.0° (p
andreyandreev [35.5K]

Answer:

The amount of potential energy that was initially stored in the spring is 88.8 J.

Explanation:

Given that,

Mass of block = 1.60 kg

Angle = 30.0°

Distance = 6.55 m

Speed = 7.50 m/s

Coefficient of kinetic friction = 0.50

We need to calculate the amount of potential energy

Using formula of conservation of energy between point A and B

U_{A}+k_{A}+w_{A}=U_{B}+k_{B}

U_{A}+0-fd=mgy+\dfrac{1}{2}mv^2

U_{A}=\mu mg\cos\theta\times d+mg h\sin\theta+\dfrac{1}{2}mv^2

Put the value into the formula

U_{A}=0.50\times1.60\times9.8\cos30\times6.55+1.60\times9.8\times6.55\sin30+\dfrac{1}{2}\times1.60\times(7.50)^2

U_{A}=88.8\ J

Hence, The amount of potential energy that was initially stored in the spring is 88.8 J.

7 0
3 years ago
For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this
ikadub [295]

Answer:

no idea

Explanation:

7 0
3 years ago
A meterstick is placed on a pivot point of 42.5cm and a 45g mass is hung at the 20cm mark. When released the meterstick remains
Vikki [24]
Hey i dont have an answer but i need the points for finals today. Thank you
6 0
3 years ago
Read 2 more answers
An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in
Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

T {V}^{\gamma - 1} = \text{constant}

\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}

Given:-

{T}_{1} = 275 \; K  

{T}_{2} = T \left( \text{say} \right)

{V}_{1}  = V

{V}_{2} = 6V

\gamma = \cfrac{5}{3} \;    (the gas is monoatomic)

\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}

 

\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}  

T  =  275 \times 0.30

T  =  83 K.

3 0
3 years ago
Calculate the intensity of current flowing through a computer that consumes 180W and operates at 120 V.
padilas [110]

Answer:

C) 1.5 A

Explanation:

P = IV

180 W = I (120 V)

I = 1.5 A

5 0
3 years ago
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