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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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Which object would have the most momentum?
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B

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Since all options have the mass and speed in the same units, there is no need for conversion.

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Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such
Lyrx [107]

This question is incomplete, the complete question is;

- Calculate the difference in blood pressure between the feet and top of the head of a person who is 1.80m Tall

- Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head

Answer:

- the difference in blood pressure is 18698.4 Pa

- the additional outward force F is 4.86 N

Explanation:

Given the data in the question;

we know that the expression for difference in blood pressure is;

ΔP = pgh

where p is density = 1060 kg/m³

g is acceleration due to gravity  = 9.8 m/s²

and h is height = 1.80 m

now we substitute

ΔP = 1060 × 9.8 × 1.80

ΔP = 18698.4 Pa

therefore the difference in blood pressure is 18698.4 Pa

Also given that;

diameter of blood vessel d = 3.10 mm

radius r = 3.10 mm / 2 = 1.55 mm = 0.00155 m

length l = 2.70 cm = 0.027 m

Surface area of the cylindrical segment of a blood vessel is

A = 2πrl

we substitute

A = 2 × π × 0.00155 × 0.027

A = 2.6 × 10⁻⁴ m²

so

the required for will be;

F = PA

we substitute

F = 18698.4 Pa × 2.6 × 10⁻⁴ m²

F = 4.86 N

Therefore, the additional outward force F is 4.86 N

5 0
3 years ago
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