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3 years ago

The statement that "The universe is the same everywhere and does not change over time" is known as _____.

Physics

2 answers:

kondor19780726 [428]3 years ago

8 0

The answer is B- Perfect Cosmological Principle

Sorry it took people so long to anser:)

djyliett [7]3 years ago

4 0

Answer: b. the perfect cosmological principle

Explanation:

The perfect cosmological principle is the extension of the cosmological principle. It states that the universe is homogeneous. It looks same from everywhere. It will look the same. It will never change over time instead of some of the gross changes occurring in the universe.

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X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{

dedylja [7]

<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$ :

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$ (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$ (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

$E_{o}=\frac{h.c}{\lambda_{o}}$ (4)

From this equation (4) we can find the value of $\lambda_{o}$ :

$\lambda_{o}=\frac{h.c}{E_{o}}$ (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$

$\lambda_{o}=3.102(10)^{-12}m$ (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$ , let's find $\lambda'$ :

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$ (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$

$\lambda'=3.427(10)^{-12}m$ (8)

Knowing the wavelength of the scattered photon $\lambda'$ , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$ (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$

$E'=362.063keV$ (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$ , we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$ (11)

$E_{e}=400keV-362.063keV$

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$

5 0

3 years ago

Buildings

Natasha2012 [34]

Buildings can affect the surface albedo.

The surface is the outermost or uppermost layer of a physical item or area. it is the element or place of the object which could first be perceived by an observer through the usage of the senses of sight and contact and is the component with which other materials first engage.

A surface is the outermost layer of any physical object. it is the part of the object that may be perceived by an observer using their experience of sight and touch and is the portion with which surrounding substances first engage.

The chemical and mechanical residences include chemical composition, grain, hardness, energy, and inhomogeneities.

Learn more about surface here:-brainly.com/question/16519513

#SPJ4

8 0

1 year ago

Does the eruption in the video constitute a pyroclastic flow

densk [106]

If it is the video I am thinking of then Yes

7 0

3 years ago

When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of

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