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2 years ago

How is the difference between science and engineering Best stated?

1 answer:

8 0

Answer: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

Explanation:

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TOPO NÀO CÓ CẤU HÌNH ĐA ĐIỂM

bixtya [17]

Answer:

Tout à fait les gens sont nuls

3 0

3 years ago

Read 2 more answers

2. One of the many methods used for drying air is to cool the air below the dew point so that condensation or freezing of the mo

REY [17]

Answer:

0.5°c

Explanation:

Humidity ratio by mass can be expressed as

the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air

Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.

Humidity ratio expressed by mass:

x = mw / ma (1)

where

x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)

mw = mass of water vapor (kg, lb)

ma = mass of dry air (kg, lb)

It can be as:

x = 0.005 (100) / [(100 - 100)]

x = 0.005 x 100 / (100 - 100)

x = 0.005 x 100 / 0

x = 0.5°c

So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c

6 0

3 years ago

A car is about to start but it blows up. what is the problem with the car<br> ?

ratelena [41]

Answer:

because there is a bomb

6 0

3 years ago

Read 2 more answers

Un material determinado tiene un espesor de 30 cm y una conductividad térmica (K) de 0,04 w/m°C. En un instante dado la distribu

aksik [14]

Answer:

Para x=0:

$\phi=1.2 W/m^{2}$

Para x=30 cm:

$\phi=-2.4 W/m^{2}$

Explanation

Podemos utilizar la ley de Fourier par determinar el flujo de calor:

$\phi=-k\frac{dT}{dx}$ (1)

Por lo tanto debemos encontrar la derivada de T(x) con respecto a x primero.

Usando la ley de potencia para la derivda, tenemos:

$\frac{dT(x)}{dx}=300x-30$

Remplezando esta derivada en (1):

$\phi=-0.04(300x-30)$

Para x=0:

$\phi=0.04(30)$

$\phi=1.2 W/m^{2}$

Para x=30 cm:

$\phi=-0.04(300*0.3-30)$

$\phi=-2.4 W/m^{2}$

Espero que te haya ayudado!

4 0

3 years ago

An Otto cycle engine is analyzed using the air standard method. Given the conditions at state 1, compression ratio (r), and pres

My name is Ann [436]

Answer:

A) 222.58 kJ / kg

B) 0.8897 M^3/ kg

c) 0.7737 m^3/kg

D) 746.542 k

E) 536.017 kj/kg

efficiency = 58% ( approximately )

Explanation:

Given Data :

Gas constant (R) = 0.287 kJ/ kg.K

T1 = 310 k

P1 ( Kpa ) = 100

r = 11.5 ( compression ratio )

rp = 1.95 ( pressure ratio )

A ) specific internal energy at state 1

= Cv*T1 = 0.718 * 310 = 222.58 kJ / kg

B) Relative specific volume at state 1

= P1*V1 = R*T1 ( ideal gas equation )

V1 = R*T1 / P1 = (0.287* 10^3*310 ) / 100 * 10^3

V1 = 88.97 / 100 = 0.8897 M^3/ kg

C ) relative specific volume at state 2

Applying r ( compression ratio) = V1 / V2

11.5 = 0.8897 / V2

V2 = 0.8897 / 11.5 = 0.7737 m^3/kg

D) The temperature (k) at state 2

since the process is an Isentropic process we will apply the p-v-t relation

$\frac{T1}{T2} = (\frac{V1}{V2}^{n-1} ) = (\frac{P2}{P1} )^{\frac{n-1}{n} }$

hence T2 = $9^{1.4-1} * 310$ = 2.4082 * 310 = 746.542 k

e) specific internal energy at state 2

= Cv*T2 = 0.718 * 746.542 = 536.017 kj/kg

efficiency = output /input = 390.3511 / 667.5448 ≈ 58%

attached is a free hand diagram of an Otto cycle is attached below

3 0

3 years ago