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3 years ago

A paragraph that can serve the specific purpose of creating a special effect within the essay is called?

Physics

2 answers:

Katarina [22]3 years ago

7 0

Its b.functional paragraph because writers use this for interest presents and special effects

fomenos3 years ago

3 0

Answer:

b. functional paragraph

Explanation:

A paragraph is a communicative unit formed by a set of sequential sentences that deals with the same topic. It is composed of a set of sentences that have a certain thematic unit or that, without having it, are enunciated together. It is a component of the text that in its external aspect begins with a capital letter and ends in a separate point. It includes several related sentences on the same sub-theme; One of them expresses the main idea.

<u>Functional paragraph</u>

Within the formal structure of the text, they play a role not so much in content development, but rather in keeping the informational fabric organized and related to each other. They are usually short paragraphs that help guide the exposition of thought, relating some paragraphs to others to contribute to unity, cohesion and coherence of the text. These paragraphs do not have a thematic sentence that expresses a central idea, since their prayers only collaborate with the development of the main propositions contained in the informative paragraphs that precede or follow them. They can be of three kinds:

Heading or introductory paragraphs: As the name implies, its function is to introduce or present the topic. You must also locate the reader and arouse his interest.

Link or transition paragraphs: They relate the information in one paragraph with that in another. Some are retrospective (when referring to information already presented) and others, prospective (when they announce information that will appear later). They can be considered as major connectors that contribute to the unity, cohesion and coherence of the text.

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Define limitations in the operation conditions of a pn junction<br>

suter [353]

Answer:

Such limitations are given below.

Explanation:

Each pn junction provides limited measurements of maximum forwarding current, highest possible inversion voltage as well as the maximum output level.
If controlled within certain adsorption conditions, the pn junction could very well offer satisfying performance.
In connector operation, the maximum inversion voltage seems to be of significant importance.

6 0

4 years ago

As the wavelength increases, the frequency (2 points) decreases and energy decreases. increases and energy increases. decreases

frozen [14]

Bohr's equation for the change in energy is
$\Delta E= \frac{hc}{\lambda}$
where
h = Planck's constant
c == the velocity of light
λ = wavelength.

The velocity is related to wavelength and frequency, f, by
c = fλ

Let us examine the given answers on the basis of the given equations.

a. As λ increases, f decreases and ΔE decreases.
TRUE

b. As λ increases, f increases and ΔE increases.
FALSE

c. As λ increases, f increases and ΔE decreases.
FALSE

Answer:
As the wavelength increases, the frequency decreases and energy decreases.

3 0

3 years ago

Read 2 more answers

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,

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