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4 years ago

Explain the benefits of reading A vocabulary section at the beginning of a text

Engineering

2 answers:

torisob [31]4 years ago

4 0

Answer:

My response: The benefits of reading a vocabulary section at the beginning of a text is so that when you are reading you ether know what the word means while your reading and you come across that word or so when you are reading and you come across that word you can use context clues to figure out what the word means.

Sample Response: A vocabulary section at the beginning helps you learn new words to look out for when you read. It tells what kind of information will be in the text. It also gives you a starting advantage to understand scientific language.

Explanation: This is what i put for mine but i will provide the sample response as well. Hope this helps!!

Zinaida [17]4 years ago

3 0

Answer: Deatiled information

Explanation:

It helps the reader organize and sort information in the text. It provides detailed information about the content of the text. It prepares readers to look for new vocabulary as they read.

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If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam

Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

$\zeta =\dfrac{C}{C_c}$

Where C is the damping coefficient and Cc is the critical damping coefficient.

$C_c=2\sqrt{mK}$

So from above we can say that

$\zeta =\dfrac{C}{2\sqrt{mK}}$

$\zeta \alpha \dfrac{1}{\sqrt K}$

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0

3 years ago

Something that pollutes a system is called a(n) ??? .

frutty [35]

Answer:

Trash pollution

correct me if wrong

Explanation:

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3 years ago

One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter

lukranit [14]

The pressure of water is 7.3851 kPa

Explanation:

Given data,

V = 150× $10^{-3}$ $m^{3}$

m = 1 Kg

$P_{1}$ = 2 MPa

$T_{2}$ = 40°C

The waters specific volume is calculated:

$v_{1}$ = V/m

Here, the waters specific volume at initial condition is $v_{1}$ , the containers volume is V, waters mass is m.

$v_{1}$ = 150× $10^{-3}$ $m^{3}$ /1

$v_{1}$ = 0.15 $m^{3}$ / Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 $m^{3}$ / Kg and 0.13 $m^{3}$ / Kg.

$T_{1}$ = 350+(400-350) $\frac{0.15-0.13}{0.1522-0.1386}$