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3 years ago

12

A benefit to using the medium the author used in "Great Rock and Roll

2 answers:

lana66690 [7]3 years ago

7 0

Incomplete question. The options read;

A. can change the story's ending

B. listens to the dialogue

C. hears the rock songs

D. feels more connected to the text.

Answer:

<u>D. feels more connected to the text.</u>

Explanation:

<em>Remember</em>, the underlying motive of an author when writing any text is to make his readers or audience more connected to the material been read or heard.

Hence, in<em> "Great Rock and Roll Pauses" </em>we can conclude that<em> </em>the author's choice of medium was motivated by the same desire of making the audience feel more connected to the text.

IgorLugansk [536]3 years ago

6 0

Answer:

The audience can actually hear the music.

Explanation: dont listen to below average iq people :0

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) Assuming different AM regulations; the receiver is using mixer with subtracting format. The frequency selectivity ratio is app

Zarrin [17]

Answer:

$F=710KHZ$

Explanation:

From the question we are told that:

Frequency selectivity ratio $R=10$

AM range 750 kHz to 2600 kHz

Therefore Bandwidth is

$B=2100-680$

$B=1420KHZ$

Generally the equation for The intermediate frequency is mathematically given by

Intermediate frequency=\frac{Bandwidth}{2}

$F=\frac{B}{2}$

$F=\frac{1420}{2}$

$F=710KHZ$

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3 years ago

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3 years ago

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The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

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Alex_Xolod [135]

Sorry need.points I'm new

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3 years ago

Affordability is most concerned with:

leva [86]

Answer:

feasibility study

Explanation:

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3 years ago