Can someone help me with this physics question? I don't know which answer is correct.

How do scientists use the Doppler effect to understand the universe?

professor190 [17]

There's a very subtle thing going on here, one that could blow your mind.

Wherever we look in the universe, no matter what direction we look,
we see the light from distant galaxies arriving at our telescopes with
longer wavelengths than the light SHOULD have.

The only way we know of right now that can cause light waves to get
longer after they leave the source is motion of the source away from
the observer. The lengthening of the waves on account of that motion
is called the Doppler effect. (The answer to the question is choice-c.)

But that may not be the only way that light waves can get stretched. It's
the only way we know of so far, and so we say that the distant galaxies
are all moving away from us.

From that, we say the whole universe is expanding, and that right there is
one of the strongest observations that we explain with the Big Bang theory
of creation.

Now: If ... say tomorrow ... a competent Physicist discovers another way
for light waves to get stretched after they leave the source, then the whole
"expanding universe" idea is out the window, and probably the Big Bang
theory along with it !

Now that our mind has been blown, come back down to Earth with me,
and I'll give you something else to think about:

It's true that when we look at distant galaxies, we do see their light
arriving in our telescopes with longer wavelengths than it should have.
And then we use the Doppler effect to calculate how fast that galaxy
is moving away from us. That's all true. Astronomers are doing it
every day. I mean every night.

So here's the question for you to think about ... maybe even READ about:

When the light from a distant galaxy pours into our telescope, and we
look at it, and we measure its wavelength, and we find that the wavelength
is longer than it should be ... how do we know what it should be ? ? ?

6 0

3 years ago

Read 2 more answers

A woman holds a book by placing it between her hands such that she presses at right angles to the front and back covers. The boo

In-s [12.5K]

Answer:

Explanation:

In order to solve this problem we need to make a free body diagram of the book and the forces that interact on it. In the picture below you can see the free body diagram with these forces.

The person holding the book is compressing it with his hands, thus exerting a couple of forces of equal magnitude and opposite direction with value F.

Now the key to solving this problem is to analyze the equilibrium condition (Newton's third law) on the x & y axes.

To find the weight of the book we simply multiply the mass of the book by gravity.

W = m*g

W = 1.3[kg] * 9.81[m/s^2]

W = 12.75 [N]

7 0

3 years ago

A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight

photoshop1234 [79]

Answer:

$t=1.9 sec$

Explanation:

From the question we are told that:

Height $h=28m$

Time $t=3s$

Generally the Newton's equation for Initial velocity upward is mathematically given by

$s=ut+\frtac{1}{2}at^2$

$28=3u-\frac{1}{2}*9.8*3^2$

$u=24.03m/s$

Generally the velocity at elevation and depression occurs as ball arrives and passes through S=28

$v=\sqrt{24.03-2*9.8*28}$

$v=5.35m/s and -5.35m/s$

Generally the Newton's equation for time to reach initial velocity is mathematically given by

$v=u+at$

$5.35=24.03-9.8t$

$t=\frac{28.03-5.35}{9.8}$

$t=1.9 sec$

4 0

3 years ago

Which activity is best described as a scientific endeavor?

Kipish [7]

Answer:

B

Explanation:

BECAUSE TO DO THE TESTS YOU NEED TO DO THE SCIENTIFIC METHOD.

FOR EXAMPLE: OBSERVATIONS AND EXPERIMENTS TO OBTAIN RESULTS.

ANYWAY I LEAVE YOU THE LINK:

https://gscourses.thinkific.com

8 0

3 years ago

Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy

Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

$P_{1}V_{1}=nRT$

$V_{1}=\dfrac{nRT}{P_{1}}$

Put the value into the formula

$V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}$

$V_{1}=2.62\times10^{-3}\ m^3$

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

6 0

3 years ago