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Gwar [14]
3 years ago
12

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

Engineering
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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A crankcase heater is often used to prevent refrigerant from mixing with compressor oil during periods of:
stira [4]

Answer:

Low ambient temperature

Explanation:

Hope this helps. If it did, please mark as brianliest so other people see it. Thanks! - Kai

5 0
2 years ago
Read 2 more answers
Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0
3 years ago
How to draw the output voltage waveform rectifier
tatyana61 [14]

Answer:

Half-wave rectifier converts an AC signal into a DC signal. It's called a half-wave because it only rectify the positive part of an AC signal.

AC Signal = An electrical signal that alternates between positive and negative voltage.

DC Signal = An electrical signal that only has positive voltage.

Rectify = A fancy word for converting something.

Adding a capacitor helps the positive part of the signal stay on longer. This work because the capacitor stores energy kinda like a battery. During the negative part of the AC signal, the energy stored in the capacitor will be drained and used, then the cycle repeats.

The load resistor is just there to prevent a short circuit from happening.

7 0
2 years ago
Technician A says mismatching tires of the same size on a heavy vehicle will generally not affect ABS operation. Technician B sa
marysya [2.9K]
Technician A is correct
3 0
3 years ago
A Pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3200
coldgirl [10]

Answer:

Speed of aircraft ; (V_1) = 83.9 m/s

Explanation:

The height at which aircraft is flying = 3000 m

The differential pressure = 3200 N/m²

From the table i attached, the density of air at 3000 m altitude is; ρ = 0.909 kg/m3

Now, we will solve this question under the assumption that the air flow is steady, incompressible and irrotational with negligible frictional and wind effects.

Thus, let's apply the Bernoulli equation :

P1/ρg + (V_1)²/2g + z1 = P2/ρg + (V_2)²/2g + z2

Now, neglecting head difference due to high altitude i.e ( z1=z2 ) and V2 =0 at stagnation point.

We'll obtain ;

P1/ρg + (V_1)²/2g = P2/ρg

Let's make V_1 the subject;

(V_1)² = 2(P1 - P2)/ρ

(V_1) = √(2(P1 - P2)/ρ)

P1 - P2 is the differential pressure and has a value of 3200 N/m² from the question

Thus,

(V_1) = √(2 x 3200)/0.909)

(V_1) = 83.9 m/s

4 0
3 years ago
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