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2 years ago

Why is the reflection step in the engineering process the most important step?

1 answer:

6 0

Reflection helps designers to learn from their experiences, to integrate and co-ordinate different aspects of a design situation, to judge the progress of the design process, to evaluate interactions with the design context, and to plan suitable future design activities.

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Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

Elodia [21]

Answer:

1.25 cm/day

Explanation:

An air thickness , (l) = 0.15 cm

Air Temperature =

$(T_a)=20^0C = (20+273)K\\(T_a)=293K$

Mass Diffusion coefficient (D) = $0.25cm^2/sec$

If the air pressure $(P_a) = 0.5 P_{sat}$

We are to determine how fast will the water $(H_2O)$ level drop in a day.

From the property of air at T = 20° C

$P_{sat} = 2.34$ from saturated water properties.

The mass flow of $(H_2O)$ can be calculated as:

$H_2O = \frac{D}{\phi} \delta C$

where:

$\delta C = \frac{P_{sat}*P_a}{RT }$

R(constant) = 8.314 kJ/mol.K

$\delta C = \frac{2.34*0.5}{8.314*293 }$

$\delta C = 4.803*10^{-4}\\ \delta C =0.48*10^{-3} mol/m^3\\ \delta C = 0.48*10^{-6} mol/cm^3$

Since 1 mole = 18 cm ³ of water

$0.48*10^{-6}mol/cm^3$ will be: $(0.48*10^{-6}mol/cm^3 *18)cm^3/cm^3$

$\delta C = 8.64 * 10^{-6}$

Again:

$H_2O = \frac{D}{\phi} \delta C$

$= \frac{0.25}{0.15}*8.69*10^{-6} \frac{cm^2/sec}{cm}$

$=1.4481*10^{-5} \frac{cm^2/sec}{cm}$

Converting the above value to cm/day: we have:

$1.448*10^{-5}*3600*24\frac{cm}{s}*\frac{s}{yr}*\frac{yr}{day}$

= 1.25 cm/day

∴ the rate at which the water level drop in a day = 1.25 cm/day

7 0

3 years ago

A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled

Vikentia [17]

Answer:

Force per unit plate area is 0.1344 $N/m^{2}$

Solution:

As per the question:

The spacing between each wall and the plate, d = 10 mm = 0.01 m

Absolute viscosity of the liquid, $\mu =1.92\times 10^{- 3} Pa-s$

Speed, v = 35 mm/s = 0.035 m/s

Now,

Suppose the drag force that exist between each wall and plate is F and F' respectively:

Net Drag Force = F' + F''

$F = \tau A$

where

$\tau$ = shear stress

A = Cross - sectional Area

Therefore,

Net Drag Force, F = $(\tau ' +\tau '')A$

$\frac{F}{A} = \tau ' +\tau ''$

Also

F = $\frac{\mu v}{d}$

where

$\mu$ = dynamic coefficient of viscosity

Pressure, P = $\frac{F}{A}$

Therefore,

$\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}$

$\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}$

8 0

2 years ago

If the electric field just outside a thin conducting sheet is equal to 1.5 N/C, determine the surface charge density on the cond

Dennis_Churaev [7]

Answer:

The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²

Explanation:

The electric field intensity due to a thin conducting sheet is given by the following formula:

Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)

From this formula:

Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)

We have the following data:

Electric Field Intensity = 1.5 N/C

Permittivity of free space = 8.85 x 10^-12 C²/N.m²

Therefore,

Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)

<u>Surface Charge Density = 26.55 x 10^-12 C/m²</u>

Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².

7 0

3 years ago

Scientific research techniques are used to analyze the effectiveness of political advertising. False True

miv72 [106K]

Answer:

correct me if i'm wrong but i think it's false

Explanation:

5 0

2 years ago

For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking.

kirza4 [7]

Answer:

203.0160

Explanation:

Because you add then subtract then multiply buy 7 the subtract then divide then you add that to the other numbers you got than boom

7 0

1 year ago