The atomic number of magnesium is 12. This means that its nucleus must contain

Near the center of our galaxy, hydrogen gas is moving directly away from us in its orbit about a black hole. The electromagnetic

klemol [59]

To solve this problem it is necessary to apply the concept related to wavelength, specifically when the wavelength is observed from a source that is in motion to the observer.

By definition the wavelength is given defined by,

$\lambda_{obs} = \lambda_s \sqrt{\frac{1+u/c}{1-u/c}}$

Where

$\lambda_{obs}$ = Observed wavelength

$\lambda_s$ = Wavelength of the source

c = Speed of light in vacuum

u = Relative velocity of the source to the observer

According to our data we have that the wavelength emitted from the galaxy is 1875nm which is equal to the wavelength from the source, while the wavelength from the observer is $\lambda_{obs}=1945nm$

Therefore replacing in the previous equation we have,

$1945 = 1875 \sqrt{\frac{1+\frac{u}{c} }{1-\frac{u}{c} }}$

$\sqrt{\frac{1+u/c}{1-u/c}} = 1.03733$

$\frac{1+\frac{u}{c} }{1-\frac{u}{c} }=1.03733^2$

$1+\frac{u}{c} =1.03733^2*(1-\frac{u}{c} )$

Solving for u,

$1+\frac{u}{c} =1.03733^2*(1-\frac{u}{c} )$

$1+\frac{u}{c} =1.03733^2-1.03733^2(\frac{u}{c} )$

$\frac{u}{c} +1.03733^2\frac{u}{c} =1.03733^2-1$

$2.88595\frac{u}{c}=1.03733^2-1$

$\frac{u}{c} = \frac{1.03733^2-1}{2.88595}$

$u = \frac{1.03733^2-1}{2.88595}*c$

$u = 0.02635c$

Therefore the speed of the gas relative to earth is 0.02635 times the speed of light.

6 0

3 years ago

Which of the following is not a way that astronomers can find how much dark matter there is in cluster of galaxies?

Tpy6a [65]

Answer:

b. Observe the radio waves coming from all dark matter; from the strength of the radio waves from each cluster, estimate the amount of dark matter needed to produce them.

Explanation:

The universe is thought to be made up of 85% dark matters. Dark matter is called dark because it does not appear to interact with the electromagnetic field, which means it doesn't absorb, reflect or emit electromagnetic radiation, and is therefore difficult to detect. This means that option b is wrong since radio wave is an electromagnetic wave. Dark matter is a form of matter that makes up about a quarter of the total mass–energy density of the universe. Dark matter was theorized due a variety of astrophysical observations and gravitational effects that cannot be explained by accepted theories of gravity unless there were more matter in the universe than can be seen.

6 0

3 years ago

What mass of water will fill a tank that is 100.0 cm long, 50.0 cm wide, and 30.0 cm high? Express the answer in grams.

Bond [772]

$calculate\ mass\ from\ formula\ for\ density:\\\\
density=volume*mass\\
mass=\frac{density}{volume}\\\\
volume=width*length*height\\
volume=100*50*30=150000cm^3=0,15m^3\\\\
density\ of\ water=\ 999.9720 \frac{kg}{m^3} \\\\mass=0,15*999.9720=149,9958kg=149995,8grams\\\\ Answer\ is\ :\\water\ which\ weight\ is\ equal\ to\ 149995,8grams.
$

4 0

3 years ago

The amplitude of a simple harmonic oscillator will be doubled by:a) doubling only the initial speedb) doubling the initial displ

OLEGan [10]

Answer:

When both initial speed and initial displacement is doubled then amplitude will be doubled.

Explanation:

Given that :- Amplitude of simple harmonic Oscillator is doubled.

So,

Formula of Simple harmonic oscillator is $X=A\sin\ (2\pi ft +\phi)$ ...........(1)

Where X = Position in (m,cm,km.....)

A = Amplitude in (m,cm,km.....)

F = Frequency in (Hz)

T = Time in (sec.)

Ф = Phase in (rad)

For initial displacement taking t=0 we get,

Initial displacement = $A\sin(\phi)$ .................(2)

Now taking equation (1) and differentiating it w.r.t to (t) we get

$\frac{dx}{dt} = 2\pi fA\cos\ (2\pi ft+\phi)$

$V= 2\pi fA\cos\ (2\pi ft+\phi)$

taking t=0 for initial speed then we get,

Initial speed = $2\pi fA\cos\phi$ ...............(3)

observing equation (2) & (3) that the initial displacement and initial speed depends on the Amplitude of the Oscillator.

Hence,

when both initial speed and displacement is doubled then amplitude will be doubled.

4 0

4 years ago

about how much more energy is released in a 6.5 richter magnitude earthquake than in one with magnitude 5.5?

OverLord2011 [107]

Answer:

For example, an earthquake of magnitude 5.5 releases about 32 times as much energy as an earthquake measuring 4.5. Another way to look at this is that it takes about 900 magnitude 4.5 earthquakes to equal the energy released in a single 6.5 earthquake.

Explanation:

8 0

2 years ago