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Ugo [173]
3 years ago
6

Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll

. ratio. (b) Plexiglas is often used to shield high-energy beta emitters rather than lead, even though lead is a better shield against the bremsstrahlung photons. Both shields will stop the high-energy beta, so why is Plexiglas used instead of lead?
Physics
1 answer:
Andrew [12]3 years ago
4 0

Answer:

Explanation:

During an energy transfer, the collision loss for an electron can be determined by using the formula:

Q = \dfrac{4mME }{(m+M)^2}

However; from the total stopping power & power loss of the electron;

\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}

where;

Z = atomic no. for lead = 82

E = 1.9 MeV

∴

radiational energy loss = collisional energy loss  =\dfrac{82 \times 1.9}{800}

= 0.19475

b)

Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.

Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.

Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.

Finally, plexiglass contains a high dimensional balance with an excessive dielectric constant.

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A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent
azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

  • <em>Length of the string, L = 100 cm</em>

<em />

The wavelengths of the constituent travelling waves is calculated as follows;

L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}

for first mode: n = 1

\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm

for second mode: n = 2

\lambda = \frac{2L}{2} = L = 100 \ cm

For the third mode: n = 3

\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm

For fourth mode: n = 4

\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50  \ cm

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

5 0
3 years ago
Can someone help me please??
Mrrafil [7]

1. Our solar system is the only place in the universe where gravity played a key part in the formation of planets.

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A psychologist is interested in exploring the effect tutorial support on students academic performance and assign students in to
NikAS [45]

Answer:

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The independent variable is the presence/absence of tutorial support

The control group are students who did not get the tutorial support.

The experimental group were students that got the tutorial support

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The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.

The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.

The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).

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<span>In the labeled portion of the curve ,you use the heat of vaporization to calculate the heat absorbed in the 4th portion. It is indicated in the picture that it is the region where vaporization occurs, that is why you need to consider this portion to calculate.</span>
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Just the answer (PLSS)
Salsk061 [2.6K]

The miracle year for Albert Einstein  was the year 1905 within which he published so many renowned papers.

<h3>When was Einstein miracle year?</h3>

The miracle year for Albert Einstein  was the year 1905 within which he published so many renowned papers in a short time and became very popular.

His mindset in that year was one that challenged the orthodox explanations and sought to think outside the box.

Learn more about Albert Einstein:brainly.com/question/2964376

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