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Delvig [45]
3 years ago
7

A star is moving away from an observer at 1% of the speed of light. At what wavelength would the observer find an emission line

which would occur at a wavelength of 6000 Angstroms if the star were at rest?
Physics
1 answer:
Ivan3 years ago
6 0

Answer:

  λ = 5940 Angstroms

Explanation:

This is an exercise of the relativistic Doppler effect

        f’= f  √((1- v / c) / (1 + v / c))

Where the speed in between the strr and the observer is positive if they move away

Let's use the relationship

         c = λ f

         f = c /λ

We replace

              c /λ’ = c /λ  √ ((1- v / c) / (1 + v / c))

              λ = λ’ √ ((1- v / c) / (1 + v / c))

Let's calculate

             v = 0.01 c

             v = 0.01 3 10⁸

             v=  3 10⁶ m / s

             λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]

             λ = 6000 √ [0.99 / 1.01]

             λ = 5940 Angstroms

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Answer:

 F= 600 N

Explanation:

Given that

Initial velocity ,u= 0 m/s

Final velocity ,v= 30 m/s

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We know that from second law of Newtons

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3 years ago
Raindrops hitting the side windows of a car in motion often leave diagonal streaks even if there is no wind. Why? Is the explana
labwork [276]

Answer:

because of the raindrop velocity relative of the car has a vertical and horizontal component  

Explanation:

  1. The car moves in a <em>horizontal direction </em>relative to the ground. The raindrops fall in the <em>vertical direction</em> relative to the ground.
  2. Their velocity relative to the moving car has  both vertical and horizontal components and this is the reason for the diagonal streaks on the side window.
  3. The diagonal streaks on the windshield  arise from a different reason.
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The number of bacteria in a certain population increases according to a continuous exponential growth model, with a growth rate
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Answer:

It would take approximately 289 hours for the population to double

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Recall the expression for the continuous exponential growth of a population:

N(t)=N_0\,e^{kt}

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In our case, we don't know No (original population, but know that we want it to double in a certain elapsed "t". We also have in mind that the percent rate "k" would be expressed in mathematical form as: 0.0024 (mathematical form of the given percent growth rate).

So we need to solve for "t" in the following equation:

2\,N_0=N_0\,e^{0.0024\,t}\\\frac{2\,N_0}{N_0} =e^{0.0024\,t}\\2=e^{0.0024\,t}\\ln(2)=0.0024\,t\\t=\frac{ln(2)}{0.0024} \\t=288.811\,\, hours

Which can be rounded to about 289 hours

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3 years ago
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