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cupoosta [38]
2 years ago
15

In the spectrum of a galaxy that is travelling away from Earth a hydrogen line is observed to be shifted from 394 nm to 450 nm.

Calculate the velocity of recession of the galaxy.
Physics
1 answer:
Murljashka [212]2 years ago
3 0

The velocity of recession of the galaxy will be 37.3×10⁶ m/sec. The concept of Doppler effect is employed in the given problem.

<h3>What is the Doppler effect?</h3>

A sudden change in the frequency due to the distance between the objects and source is explained by the Doppler effect.

As the source and observer travel toward each other, the frequency of sound, light, or other waves increases or decreases.

A passing siren's rapid change in pitch, as well as astronomers' redshift, are both caused by this phenomenon.

Applications of the Doppler effect;

The rest wavelength is,\rm \lambda_0 = 394 \  nm

The distant wavelength is,\rm \lambda = 450 \ nm

The velocity of recession  is,V

The recession velocity is found as;

\rm V = (1-\frac{\lambda}{\lambda_0} )C \\\\ \rm V = (1-\frac{394}{450} )\times 3 \times 10^8  \\\\ V = (1- 0.875)\times 3 \times 10^8 \\\\ V = 37.3  \times 10^6  \ m/sec

Hence, the velocity of recession of the galaxy will be 37.3×10⁶ m/sec.

To learn more about the Doppler effect, refer to the link;

brainly.com/question/15318474

#SPJ1

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What type of material is good at transferring heat?
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Materials that are good conductors of thermal energy are called thermal conductors. Metals are very good thermal conductors. Materials that are poor conductors of thermal energy are called thermal insulators. Gases such as air and materials such as plastic and wood are thermal insulators
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Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres
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According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )

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Given, v_{1} =1.4 m/s

Now from equation continuity,

v_{1} A_{1} = v_{2} A_{2}.

Here, A_{1} and A_{2} are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}.

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}

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Substituting these values  with the density of water is 1000 \ kg/m^3 in pressure difference formula we get.

\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa

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Four solid plastic cylinders all have radius 2.41 cm and length 5.94 cm. Find the charge of each cylinder given the following ad
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Answer:

Check explanation

Explanation:

QUICK NOTE: THE QUESTION IS NOT COMPLETE. Although it is not, we can make assumptions, since we only need values for the UNIFORM CHARGE DENSITY.

SO, LET US BEGIN;

To solve this question we are to use the equation (1) below;

Charge,Q = uniform charge density,p × Total area of the cylinder,A ------------------------------------------------------------------------(1).

From the question, we are given radius, R to be 2.41 cm and length, L to be 5.94 cm.

Step one: calculate for the total area of the cylinder, A.

Total area of the cylinder, A= area of the top surface + area of the buttom + area of the curved surface of the cylinder.

Hence, total area of the cylinder,A is;

==> πR^2 + πR^2 + 2πRL. -------------------------------------------------------------------------(2).

Then, total area of the cylinder,A is;

==> (L + R)2πR.

Step two: find the charge of each cylinder.

===> For the first cylinder; we have the uniform charge density to be 35 nC/m^2.

Therefore, the combination of equation (1) and (3) gives;

Charge Q= p × (L + R)2πR...----------------------------(4)

Hence, Q= 35 × [(5.94 + 2.41) 2× 3.143 × 5.94].= 10912.615 coulumb.

====> For the second cylinder, we have a uniform charge density of 50 nC/m^2.

Using equation (4), charge,Q= 15,589.45 Coulumb

=====> For THE third cylinder, the uniform charge density is 600, we make use of equation (4);

Charge,Q= 600×311.789.

Charge,Q= 187,073.4 coulumb.

====> For THE fourth cylinder, the uniform charge density is 750 nC/m^2.., we make use of equation (4);

Charge,Q= 233,841.75 coulumb.

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Answer:

True

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