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3 years ago

7

I'm so confused to this question What is the difference between Science and Engineering?

2 answers:

madreJ [45]3 years ago

5 0

Answer:

<u>Generally, Science is the study of the physical world, while Engineering applies scientific knowledge to design processes, structures or equipment.</u>

muminat3 years ago

4 0

Explanation:

Science is the body of knowledge that explores the physical and natural world. Engineering is the application of knowledge in order to design, build and maintain a product or a process that solves a problem and fulfills a need (i.e. a technology).

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The displacement volume of an internal combustion engine is 2.2 liters. The processes within each cylinder of the engine are mod

soldier1979 [14.2K]

Answer:

A) 14.75

B) 3.36Kj

C) 2384.2k

D) 117.6kW

E) 57.69%

Explanation:

Attached is the full solutions.

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3 years ago

if when you put your shirt in your pants, your shirt is tucked, does that mean when your shirt is over your pants, your pants ar

Archy [21]

Answer:

confusing, but yes

Explanation:

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2 years ago

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almond37 [142]

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3 years ago

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A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. (a) If the co

podryga [215]

Answer:

a) $T_{H} = 1.967\,^{\circ}F$ , b) $COP_{R} = 9.105$ , c) $T_{H} = 115.934\,^{\circ}F$ , d) $COP_{R} = 6.995$ , e) $T_{H} = 25.129\,^{\circ}C$

Explanation:

a) The coefficient of performance of the reversible refrigeration cycle is:

$COP_{R} = \frac{T_{C}}{T_{H}-T_{C}}$

$10 = \frac{419.67\,R}{T_{H}-419.67\,R}$

The temperature of the hot reservoir is:

$10\cdot T_{H} - 4196.7 = 419.67$

$T_{H} = 461.637\,R$

$T_{H} = 1.967\,^{\circ}F$

b) The coefficient of performance is:

$COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}$

$COP_{R} = 9.105$

c) The temperature of the hot reservoir can be determined with the help of the following relation:

$\frac{Q_{C}}{Q_{H}-Q_{C}} = \frac{T_{C}}{T_{H}-T_{C}}$

$\frac{500\,BTU}{600\,BTU-500\,BTU} = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5 = \frac{479.67\,R}{T_{H}-479.67\,R}$

$5\cdot T_{H} - 2398.35 = 479.67$

$T_{H} = 575.604\,R$

$T_{H} = 115.934\,^{\circ}F$

d) The coefficient of performance is:

$COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}$

$COP_{R} = 6.995$

e) The temperature of the cold reservoir is:

$8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}$

$8.9\cdot T_{H} - 2386.535 = 268.15$

$T_{H} = 298.279\,K$

$T_{H} = 25.129\,^{\circ}C$

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2 years ago

If the load parameters are: Vln=600kV, Il=100A (resistive), calculate the source voltage and current when the line is 50Miles (s

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Explanation:

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3 years ago