Mong m.n giúp mình vs ạ

In a morphological matrix, which of the following contains the parameters that are essential to a design?

choli [55]

In a morphological matrix, the parameters that are essential for a design are in the left column.

<h3>What is a morphological matrix?</h3>

The morphological matrix is a matrix where columns and rows represent the various parameters for solving a problem. The first column is used for the characteristics relevant to the problem; the horizontal lines are filled with possibilities for each of these parameters.

With this information, we can conclude that in a morphological matrix, the parameters that are essential for a project are in the left column.

Learn more about morphological matrix in brainly.com/question/21120930

8 0

3 years ago

What is Euler's equation?

igomit [66]

Answer:

$e^{ix} = cosx + i sinx$

Explanation:

In mathematics, Euler's formula is an equation in complex analysis, that gives a relationship between an exponential factor and the trigonometric functions.

The Euler equation is:

$e^{ix} = cosx + i sinx$

Here,

e - base of the natural logarithm

i - imaginary unit

x - argument given in radians

sin , cos - trigonometric functions sine and cosine respectively.

4 0

4 years ago

Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cro

Liula [17]

Answer:

The answer is "0.0728"

Explanation:

Given value:

$P_0= 14.696\ ps\\\\\ p _{0}= 0.00238 \frac{slue}{ft^{3}}\\\\\ A= 0.05 ft^2\\\\\ T_0= 59^{\circ}f = 518.67R\\\\\ air \ k= 1\\\\ \ cirtical \ pressure ( P^*)=P_0\times \frac{2}{k+1}^{\frac{k}{k-1}}\\$

$= 14.696\times (\frac{2}{1.4+1})^{\frac{1.4}{1.4-1}}\\\\=7.763 Psia\\\\$

if $P$ flow is chocked

if $P>P^{*} \to$ flow is not chocked

When P= 10 psia < $P^{*}$ $\to$ not chocked

match number:

$\ for \ P= \ 10\ G= \sqrt{\frac{2}{k-1}[(\frac{\ p_{0}}{p})^{\frac{k-1}{k}}-1]}$

$= \sqrt{\frac{2}{1.4-1}[(\frac{14.696}{10})^{\frac{1.4-1}{1.4}}-1]}$

$M_0=7.625$

$p=p_0(1+\frac{k-1}{2} M_0 r)^\frac{1}{1-k}$

$=0.00238(1+\frac{1.4-1}{2}0.7625`)^{\frac{1}{1-1.4}}\\\\\ p=0.001808 \frac{slug}{ft^3}$

$\ T= T_0(1+\frac{k-1}{2} Ma^r)^{-1}\\\\\ T=518.67(1+\frac{1.4-1}{2} 0.7625^2)^{-1}\\\\\ T=464.6R\\\\$

$\ velocity \ of \ sound \ (C)=\sqrt{KRT}\\\\$

$=\sqrt{1.4\times1716\times464.6}\\\\=1057 ft^3\\\\$

R= gas constant=1716

$m=PAV\\\\$

$=0.001808\times0.05\times(Ma.C)\\\\=0.001808\times0.05\times0.7625\times 1057\\\\=0.0728\frac{slug}{s}$

5 0

3 years ago

Air at 1 atm enters a thin-walled ( 5-mm diameter) long tube ( 2 m) at an inlet temperature of 100°C. A constant heat flux is ap

alexdok [17]

Answer:

heat rate = 7.38 W

Explanation:

Given Data:

Pressure = 1atm

diameter (D) = 5mm = 0.005m

length = 2

mass flow rate (m) = 140*10^-6 kg/s

Exit temperature = 160°C,

At 400K,

Dynamic viscosity (μ) = 22.87 *10^-6

Prandtl number (pr) = 0.688

Thermal conductivity (k) = 33.65 *10^-3 W/m-k

Specific heat (Cp) = 1.013kj/kg.K

Step 1: Calculating Reynolds number using the formula;

Re = 4m/πDμ

= (4*140*10^-6)/(π* 0.005*22.87 *10^-6)

= 5.6*10^-4/3.59*10^-7

= 1559.

Step 2: Calculating the thermal entry length using the formula

Le = 0.05*Re*Pr*D

Substituting, we have

Le = 0.05 * 1559 * 0.688 *0.005

Le = 0.268

Step 3: Calculate the heat transfer coefficient using the formula;

Nu = hD/k

h = Nu*k/D

Since Le is less than given length, Nusselt number (Nu) for fully developed flow and uniform surface heat flux is 4.36.

h = 4.36 * 33.65 *10^-3/0.005

h = 0.1467/0.005

h = 29.34 W/m²-k

Step 4: Calculating the surface area using the formula;

A = πDl

=π * 0.005 * 2

=0.0314 m²

Step 5: Calculating the temperature Tm

For energy balance,

Qc = Qh

Therefore,

H*A(Te-Tm) = MCp(Tm - Ti)

29.34* 0.0314(160-Tm) = 140 × 10-6* 1.013*10^3 (Tm-100)

0.921(160-Tm) = 0.14182(Tm-100)

147.36 -0.921Tm = 0.14182Tm - 14.182

1.06282Tm = 161.542

Tm = 161.542/1.06282

Tm = 151.99 K

Step 6: Calculate the rate of heat transferred using the formula

Q = H*A(Te-Tm)

= 29.34* 0.0314(160-151.99)

= 7.38 W

the Prandtl number using the formula

5 0

3 years ago

Why/how is a paperclip able to float on water?

abruzzese [7]

Answer:

Water surface tension

Explanation:

The water around the paperclip forms a kind of elastic surface, deforming, in which the clip can stay afloat.

This is because the molecules that are on the surface of the water, already in contact with the air, try to cling to those that are next to them and those that are immediately below.

If we added even if it were just a drop of soap in the water, the clip would go to the bottom, because the soap has the ability to decrease the surface tension of the water.

7 0

3 years ago