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cupoosta [38]
1 year 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]1 year 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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Consider that 168.0 J of work is done on a system and 305.6 J of heat is extracted from the system. In the sense of the first la
Ilia_Sergeevich [38]

To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.

dQ = dU + dW

Where,

Q = Heat

U = Internal Energy

By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore

W = -168J \righarrow  Work is done ON the system

Q = -305.6J \rightarrow Heat is extracted FROM the system

Therefore the value of the Work done on the system is -158.0J

3 0
3 years ago
(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down
slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

h=ut+\dfrac{1}{2}at^2

Here, u = 0 and a = g

h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s

So, it will take 2.22 s for the pumpkin to hit the ground.

7 0
2 years ago
A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju
denpristay [2]

Answer:

a) h=250\ m

b) \Delta h=0.0835\ m

Explanation:

Given:

  • upward acceleration of the helicopter, a=5\ m.s^{-2}
  • time after the takeoff after which the engine is shut off, t_a=10\ s

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

h=u.t+\frac{1}{2} a.t^2

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

h=0+0.5\times 5\times 10^2

h=250\ m

b)

  • time after which Austin Powers deploys parachute(time of free fall), t_f=7\ s
  • acceleration after deploying the parachute, a_p=2\ m.s^{-2}

<u>height fallen freely by Austin:</u>

h_f=u.t_f+\frac{1}{2} g.t_f^2

where:

u= initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

t_f= time of free fall

h_f=0+0.5\times 9.8\times 7^2

h_f=240.1\ m

<u>Velocity just before opening the parachute:</u>

v_f=u+g.t_f

v_f=0+9.8\times 7

v_f=68.6\ m.s^{-1}

<u>Time taken by the helicopter to fall:</u>

h=u.t_h+\frac{1}{2} g.t_h^2

where:

u= initial velocity of the helicopter just before it begins falling freely = 0

t_h= time taken by the helicopter to fall on ground

h= height from where it falls = 250 m

now,

250=0+0.5\times 9.8\times t_h^2

t_h=7.1429\ s

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

t'=t_h-t_f

t'=7.1428-7

t'=0.1428\ s

<u>Now the height fallen in the remaining time using parachute:</u>

h'=v_f.t'+\frac{1}{2} a_p.t'^2

h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2

h'=9.8165\ m

<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>

\Delta h=h-(h_f+h')

\Delta h=250-(240.1+9.8165)

\Delta h=0.0835\ m

5 0
2 years ago
Please can someone help me im struggling asap ill mark brainlist
juin [17]

Answer:

1. A

2. C

3. A

4. D

5. B

Explanation:

4 0
2 years ago
Read 2 more answers
It would really help if anyone could answers please and thanks
mojhsa [17]

well it would be A because 55 degrees is going strait well 75 is going literally straight up

4 0
2 years ago
Read 2 more answers
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