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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.

Functional paragraph

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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3 years ago

Read 2 more answers

Two pieces of clay are moving directly toward each other. When they collide, they stick together and move as one piece. One piec

Wittaler [7]

Answer:

The fraction of the total initial kinetic energy is lost during the collision is $\dfrac{11}{17}\ J$

Explanation:

Given that,

Mass of one piece = 300 g

Speed of one piece = 1 m/s

Mass of other piece = 600 g

Speed of other piece = 0.75 m/s

We need to calculate the final velocity

Using conservation of energy

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v$

Put the value intro the formula

$300\times10^{-3}\times1+600\times10^{-3}\times(0.75)=(300\times10^{-3}+600\times10^{-3})v$

$v=\dfrac{00\times10^{-3}\times1+600\times10^{-3}\times(-0.75)}{(300\times10^{-3}+600\times10^{-3})}$

$v=-0.5\ m/s$

We need to calculate the total initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}\times300\times10^{-3}\times1^2+\dfrac{1}{2}\times600\times10^{-3}\times(0.75)^2$

$K.E_{i}=0.31875\ J$

We need to calculate the total final kinetic energy

Using formula of kinetic energy

$K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Put the value into the formula

$K.E_{f}=\dfrac{1}{2}\times(300\times10^{-3}+600\times10^{-3})\times(-0.5)^2$

$K.E_{f}=0.1125\ J$

We need to calculate the energy lost during the collision

Using formula of energy lost

$energy\ lost=\dfrac{0.31875-0.1125}{0.31875}$

$energy\ lost=\dfrac{11}{17}\ J$

Hence, The fraction of the total initial kinetic energy is lost during the collision is $\dfrac{11}{17}\ J$

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4 years ago

What is the highest energy of radiation?

ser-zykov [4K]

Gamma rays ..........................

6 0

4 years ago

A domestic water heater holds 189 L of water at 608C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in

Gekata [30.6K]

The energy of the hot water is 482630400 J

Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,

c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C

Substituting the values of the variables into the equation, we have

Q = mcΔT

Q = ρVcΔT

Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C

Q = 482630400 J

So, the energy of the hot water is 482630400 J

The elevation the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water is 49248 m.

Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m

Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J

So, ΔU = mgΔh

ΔU = mg(h - h')

making h subject of the formula, we have

h = h' + ΔU/mg

Substituting the values of the variables into the equation, we have

h = h' + ΔU/mg

h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)

h = 0 m + 482630400 J/(9800 kgm/s²)

h = 0 m + 49248 m

h = 49248 m

So, the elevation the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water is 49248 m.

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brainly.com/question/11961649

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3 years ago

Calculate the energy of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49 × 1014 hz. calculate the

olga nikolaevna [1]

The energy of a single photon is given by:
$E=hf$
where
E is the energy
h is the Planck constant
f is the frequency of the light

The light in our problem has a frequency $f=5.49 \cdot 10^{14}Hz$ , so the energy of each photon of that light is:
$E=hf=(6.6 \cdot 10^{-34} Js)(5.49 \cdot 10^{14} Hz)=3.64 \cdot 10^{-19} J$

5 0

3 years ago