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Citrus2011 [14]
3 years ago
6

From laboratory measurements, we know that a particular spectral line formed by hydrogen appears at a wavelength of 486.1 nanome

ters (nm). The spectrum of a particular star shows the same hydrogen line appearing at a wavelength of 485.9 nm. What can we conclude
Physics
1 answer:
MArishka [77]3 years ago
3 0

Answer:

We conclude that star moving toward observer.

Explanation:

According to the theory of doppler effect of light.

There are two kind of shift 1 )  Blue shift 2 ) Red shift, in which

1 ) if source of light moving toward observer so the apparent frequency of light increase( wavelength decrease ). This is which we call blue shift

2 ) if the distance between source and observer is increase, so the apparent frequency of light decrease ( wavelength increase ). This is which we call red shift.

Here, in our case wavelength of H-star is decrease so we can say that star moving toward observer ( Blue shift ).

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Which mathematically describes the wave properties of electrons?.
Afina-wow [57]
The Quantum Theory

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8 0
3 years ago
Read 2 more answers
A 16.9 kg monkey is swinging on a 5.32 m long vine. It starts at rest, with the vine at a 43.0° angle. How fast is the monkey mo
NemiM [27]

Answer:

v = 5.7554 m/s

Explanation:

First of all we need to know if the angle of the vine is measured in the horizontal or vertical.

To do this easier, let's assume the angle is measured with the horizontal. In this case, the innitial height of the monkey will be:

h₀ = h sinα

h₀ = 5.32 sin43° = 3.6282 m

As the monkey is dropping from the innitial point which is the suspension point, is also dropping from 5.32. Then the actual height of the monkey will be:

Δh = 5.32 - 3.63 = 1.69 m

In order to calculate the speed of the monkey we need to understand that the monkey has a potential energy. This energy, because of the gravity, is converted in kinetic energy, and the value will be the same. Therefore we can say that:

Ep = Ek

From here, we can calculate the speed of the monkey.

Ep = mgΔH

Ek = 1/2 mv²

The potential energy is:

Ep = 16.9 * 9.8 * 1.69 = 279.9

Now with the kinetic energy:

1/2 * (16.9) * v² = 279.9

v² = (279.9) * 2 / 16.9

v² = 33.12

v = √33.12

<h2>v = 5.7554 m/s</h2>

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3 0
3 years ago
Need help with this!!! <br> 15 points!!!!
irinina [24]

Answer:

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var iframe = document.getElementById("stageFrame");

for (i = 0; i < 1; i++) {

 setInterval(function() {

  if (running == null) { running = true; console.log('Program Loaded! Enjoy!') }

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Explanation:

5 0
3 years ago
The trajectory of a rock ejected from the Kilauea volcano, with a velocity of magnitude 6.4 m/s and at angle 2.2 degrees above t
Vika [28.1K]

Answer:

v = 8.03 m / s

Explanation:

This is a missile throwing exercise.

          y = y₀ + v_{oy}  t - ½ g t²

     

indicates that y = -1.2 m and the initial velocity is

         v_{oy} = v₀ sin θ

         v_{oy} = 6.4 sin 2.2

         v_{oy} = 0.2457 m / s

we substitute in the equation

         -1.2 = 0.2457 t - ½ 9.8 t²

          4.9 t² - 0.2457 t - 1.2 = 0

          t² - 0.05014 t -0.2449 = 0

we solve the quadratic equation

          t = [0.05014 ±√ (0.05014² + 4 0.2449)] / 2

          t = [0.05014 ± 0.9910] / 2

          t₁ = 0.5206 s

          t₂ = -0.47 s

time must be a positive magnitude therefore the correct answer is

          t = 0.5206 s

with this time we can calculate the vertical speed when the rock hits the ground

         v_{y} = v_{oy} - gt

         v_{y} = 0.2457 - 9.8 0.5206

         v_{y} = -4.856 m / s

the negative sign indicates that the speed is down

horizontal velocity is constant, due to no acceleration

         vₓ = v₀ₓ = v₀ cos 2,2

         v₀ₓ = 6.4 cos 2.2

         v₀ₓ = 6.395 m / s

therefore let's use Pythagoras' theorem to find the velocity

         v = √ (vₓ² + v_{y}^{2})

         v = √ (6,395² + 4,856²)

         v = 8.03 m / s

the direction can be found with trigonometry

        tan θ = v_{y} / vₓ

        θ = tan⁻¹ (-4,856 / 6,395)

        θ = - 37

the negative sign indicates that it is half clockwise from the x axis

   

4 0
4 years ago
How would you differentiate the theoretical and actual velocities of PE and KE in falling objects?
Solnce55 [7]

Answer: you've been trolled

Explanation:

8 0
3 years ago
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