Help please!

Aluminium alloys find use in aircraft industry because of

SashulF [63]

Alloys of aluminium find use in aircraft industry because of its: B) low specific gravity.

<h3>What is an alloy?</h3>

An alloy simply refers to a homogenous mixture (substance) that is produced by melting or joining two or more chemical elements together, with at least one of the elements being a metal.

Generally, some examples of the components of alloys include the following:

Carbon

Zinc

Silver

Tin

Aluminum

In the aircraft industry, alloys of aluminum typically find use because of its low specific gravity.

Read more on alloy here: brainly.com/question/17153195

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8 0

2 years ago

Define volume flow rate Q of air flowing in a duct of area A with average velocity V

Shalnov [3]

Answer:

The volume flow rate of air is $Q=A\times V$

Explanation:

A random duct is shown in the below attached figure

The volume flow rate is defined as the volume of fluid that passes a section in unit amount of time

Now by definition of velocity we can see that 'v' m/s means that in 1 second the flow occupies a length of 'v' meters

From the attached figure we can see that

The volume of the prism that the flow occupies in 1 second equals

$Volume=Area\times V=A\times V$

Hence the volume flow rate is $Q=V\times A$

3 0

3 years ago

If in Example 1.2,q = (10 - 10e2) mC, find the current at t = 0.5 s.

NeTakaya

Answer:

pls mark me as brainliest

Explanation:

Answer: 7.36 mA

3 0

3 years ago

A relatively nonvolatile hydrocarbon oil contains 4.0 mol % propane and is being stripped by direct superheated steam in a strip

hram777 [196]

Answer:

Number of Trays = Six (6)

Explanation:

Given that: y' = 25x' , in terms of molecular ratio, we can write it as

$\frac{Y'}{1 + Y'} =25 \frac{X'}{1 + X'}$ ......... 1

after plotting this we get equilibrium curve as shown in the attached picture.

inlet concentration and outlet concentration of liquid phase is

x₂ = 4% = 0.04 (inlet)

so that can be converted into molar

$X_2 = \frac{x_2}{1-x_2} = \frac{0.04}{1-0.04} = 0.04167$

and

x₁ = 0.2% = 0.002

$X_1 = \frac{x_1}{1-x_1} = \frac{0.002}{1-0.002} = 2.004*10^{-3}$

Now we have to use the balance equation a

$\frac{G_s}{L_s} = \frac{X_2-X_1}{Y_2-Y_1}$ .............. a

here amount of solute is comparably lower than

Here we have

L = 300 kmol (total)

$L_s$ = 300(1 - 0.04) = 288 kmol pure oil

G = $G_s$ = 11.42 kmol

$Y_1$ = 0 , solvent free steam

substitute into the equation a

$\frac{11.42}{288} = \frac{0.04167 - 2*10^{-3}}{Y_2 - 0}$

Y₂ = 1.0003

Now plot the point A(X₁ , Y₁) and B(X₂ , Y₂) and join them to construct operating line AB.

Starting from point B, stretch horizontal line up to equilibrium curve and from there again go down to operating line as shown in the picture attached. This procedure give one count of tray and continue the same procedure up to end of operating.

at last count, the number of stage, gives 6.

∴ <em>Number of trays = 6</em>