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Scorpion4ik [409]

3 years ago

15

Hey, I have a question, I was thinking that if you have engineering skills or drawing skill you could help me to start a project

on scratch, which would be a life or WW3 simulation.
Comment or answer if you would like to help.

1 answer:

gayaneshka [121]3 years ago

7 0

Answer

Dont have one;(

Explanation:

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Define, in words the following i relative humidity ii) dew point temperature

defon

Explanation:

i) When we divide water vapor's partial pressure by the water's equilibrium vapor pressure at a given temperature we get relative humidity.

It is denoted by ∅

$\phi ={\frac {p_ {H_{2}O} }{p_{H_{2}O} ^{*}}}$

ii) When air is cooled so that it becomes saturated with water vapor the temperature where this occurs is called the dew point.

7 0

3 years ago

3–102 One of the common procedures in fitness programs is to determine the fat-to-muscle ratio of the body. This is based on the

gayaneshka [121]

Answer:

x_fat = [ 0.5*(Wsa + Wsw) - p_muscle*V ] / V*( p_fat - p_muscle )

Explanation:

Given:

- The total volume of body = V

- The average density of the body = p_avg

- The density of muscle = p_muscle

- The density of fat = p_fat

Find:

Obtain a relation for the volume fraction of body fat x_fat

Solution:

- The volume of the fat is given by:

V_fat = x_fat*V

- The volume of the muscle is given by:

V_muscle = V - V_fat

= V - x_fat*V

=V*( 1 - x_fat )

- We will use the conservation of mass for the body related as:

mass_fat + mass_muscle = Total average mass

p_fat*V_fat + p_muscle*V_muscle = p_avg*V

p_fat*x_fat*V + p_muscle*V*( 1 - x_fat ) = p_avg*V

p_fat*x_fat + p_muscle*( 1 - x_fat ) = p_avg

- To determine p_1 we weigh the body in air:

Weight reading (Wsa) = m = p_1*V

p_1 = Wsa / V*g

- To determine p_2 we weigh the body in water:

Weight reading (Wsw) = m - p_w*V= p_1*V - p_w*V

Weight reading (Wsw) = V*(p_1 - p_w) = V*(p_2)

Where, p_2 = p_1 - p_water

p_2 = Wsw / V

- The average density p_avg:

p_avg = 0.5*(p_1 + p_2)

p_avg = 0.5*(Wsa / V + Wsw / V)

p_avg = 0.5*(Wsa + Wsw) / V

- Plug in the mass equation:

p_fat*x_fat + p_muscle*( 1 - x_fat ) = 0.5*(Wsa + Wsw) / V

x_fat*( p_fat - p_muscle ) = 0.5*(Wsa + Wsw) / V - p_muscle

x_fat = [ 0.5*(Wsa + Wsw) - p_muscle*V ] / V*( p_fat - p_muscle )

6 0

3 years ago

A liquid-liquid extraction process consists of two units, a mixer and a separator. One inlet stream to the mixer consists of two

vesna_86 [32]

Answer:

One inlet stream to the mixer flows at 100.0 kg/hr and is 35wt% species-A and 65wt% species-B.

Explanation:

8 0

3 years ago

3Px=y−y2p2<br><br>first order higher dgree

dlinn [17]

Answer:

the order higher is 3p79g5t88=yv5379

7 0

2 years ago

P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame

eimsori [14]

Answer:

a) $m=0.17kg/s$

b) $Ma=0.89$

Explanation:

From the question we are told that:

Pressure $P=60kPa$

Diameter $d=3cm$

Generally at sea level

$T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4$

Generally the Power series equation for Mach number is mathematically given by

$\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}$

$\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}$

$Ma=0.89$

Therefore

Mass flow rate

$\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\rho=0.848kg/m^3$

Generally the equation for Velocity at throat is mathematically given by

$V=Ma(r*T_0\sqrt{T_e}$ )

Where

$T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}$

$T_e=248$

Therefore

$V=0.89(1.4*288\sqrt{248})\\\\V=284$

Generally the equation for Mass flow rate is mathematically given by

$m=\rho*A*V$

$m=0.84*\frac{\pi}{4}*3*10^{-2}*284$

$m=0.17kg/s$

6 0

2 years ago