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Bad White [126]
3 years ago
12

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

radiation we receive from this hydrogen gas has a Doppler-shifted wavelength of 1940 nm when it reaches earth. From experiments done on earth, we know the source wavelength is 1875 nm. What is the speed of the gas relative to earth?
Physics
1 answer:
klemol [59]3 years ago
6 0

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.

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Please help in physics
Minchanka [31]
As there is no postive or negative assigned so
Initial velocity= -2.8759
Displacement= 0.5at^2+ut
= 0.5(-1.77)(3.33)^2+(-2.8759)(3.33)=-19.4m
4 0
3 years ago
A coil with 150 turns and a cross-sectional area of 1.00 m2 experiences a magnetic field whose strength increases by 0.65T in 1.
aleksley [76]

Answer:

54.17volts

Explanation:

Induced emf in a coil placed in a magnetic field can be expressed as E = N¶/t where

N is the number of turns = 150turns

¶ is the magnetic flux = magnetic field strength (B) × area(A)

¶ = BA

B = 0.65T

A = 1.0m²

t is the time =1.8s

Substituting this value in the formula

E = NBA/t

E = 150×0.65×1.0/1.8

E = 54.17Volts

The induced emf in the coil is 54.17Volts

7 0
3 years ago
Water is traveling into what sphere during infiltration?
o-na [289]
When precipitation hits the ground and hydrosphere
5 0
3 years ago
A train travels from the bottom of a hill to the top of a hill, its moves slower by the time it reaches the top is an example of
Neko [114]
The answer is letter C. <span>A train travels from the bottom of a hill to the top of a hill, its moves slower by the time it reaches the top is an example of negative acceleration.

In physics, acceleration can be described as an objects change of velocity. When an object gains velocity, it is positive acceleration, and negative acceleration for the opposite.

</span>

Thank you for posting your question. I hope you found what you were after. Please feel free to ask me more.

<span> </span>

5 0
3 years ago
Read 2 more answers
lasie4. A 4 kg object is displaced to the right by a distance of 12 m underthe influence of the following forces: a 17 Nforce pu
Oksanka [162]

The work done by a constant force in a rectilinear motion is given by:

W=Fd\cos\theta

where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.

In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:

\begin{gathered} W=(17)(12)\cos0+(36)(12)\cos30 \\ W=578.123 \end{gathered}

Therefore, the total work done is 578.123 J and the answer is option E

6 0
1 year ago
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