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Artemon [7]
3 years ago
9

Star #1 is approaching the Earth with speed v. Star #2 is receding from the Earth with the same speed v. Measurements of the sam

e spectral line from each of the stars show Doppler shifts in frequency. The light from which star will have the larger magnitude shift in frequency? a. star #1 b. Both stars will have the same shift. c. The value of the speed must be known before an answer can be found. d. star #2
Physics
1 answer:
leva [86]3 years ago
4 0

Answer:

b. Both stars will have the same shift.

Explanation:

It's a very simple problem to solve. Star 1 is approaching toward Earth with a speed v, so let's assume that the change in Doppler Shift is +F and Star 2 is moving away so the change in Doppler shift is -F. But it's time to notice the speed of both stars and that is same but only directions are different. speed is the main factor here. The magnitude of both shifts is F as we can see and + and - are showing there direction of motion. So, because of same amount of speed, both stars will have same shift magnitude. (Just the directions are different)

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A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i
vovangra [49]

Answer:

c=71.4m/s

\theta=8.54\textdegree

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed V_w= 15 m/s \angle  45 \textdegree

Generally Resolving vector mathematically

  sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6

Generally the equation Pythagoras theorem is given mathematically by

c^2=a^2+b^2

c^2=10.6^2 +(10.6+60)^2

c=\sqrt{10.6^2 +(10.6+60)^2}

Therefore Resultant velocity (m/s)

c=71.4m/s

b)Resultant direction

Generally the equation for solving Resultant direction

\theta=tan^-1(\frac{y}{x})

Therefore

\theta=tan^-1(\frac{10.6}{70.6})

\theta=8.54\textdegree

7 0
2 years ago
A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda  is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta  is the angle of scattering.

Given that, the scattering angle is, \theta=157^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8}  } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

here, \vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

This is the de Broglie wavelength of the electron after scattering.

8 0
3 years ago
You need to determine the density of a ceramic statue. if you suspend it from a spring scale, the scale reads 32.4 n . if you th
Mandarinka [93]
Well, the density of the water is
{1000 \frac{kg}{ {m}^{3} } }
so i believe that is what the question is asking for :)
4 0
3 years ago
Monochromatic light of wavelength, lambda, is traveling in air. The light then strikes a thin film having an index of refraction
kumpel [21]

Answer:

The  correct option is  H

Explanation:

From the question we are told that

    The index of refraction of  coating is  n_1

       The  index of refraction of material  is  n_2

   

Generally the condition for constructive for a thin film interference is mathematically represented

            2 *  t  = [ m  + \frac{1}{2}] \frac{\lambda}{n_1 }

Here  t represents the thickness

For minimum thickness  m =  0

So

           2 *  t  =0  + \frac{1}{2}\frac{\lambda}{n_1 }

=>        t  =\frac{\lambda}{4n_1 }

3 0
3 years ago
In a home's central heating system, when the room temperature measured by the thermostat reaches a set value, the thermostat sen
shutvik [7]
These actions are an example of feedback.
Given that the room has reached the desired temperature, there is no more need for it to be heated, at least until the temperature drops a bit. This is why the thermostat sends feedback about this situation to the heater, which immediately switches off until it is needed again.
3 0
3 years ago
Read 2 more answers
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