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4 years ago

Name two types of fabric testers and explain how they work.

Engineering

1 answer:

Lilit [14]4 years ago

4 0

Answer:

Merchandisers are the key persons in apparel business. They work as a bridge between buyer and seller. Merchandisers have to do various tasks from order execution to final dispatch. Developing Lab dip is one of the primary jobs for a merchandiser to start an order execution. They need to run for lab dip approval. Though this would be an easy task, sometimes big mistakes may occur. They have to do different fabric tests as buyer required. In this article I will briefly described different types of fabric test are completed by garment merchandisers.

It is necessary to validate the information that the mill or manufacturer gives you. This is normally done by having fabric tested by an accredited testing house (internationally recognized) either in the country of manufacturer or locally in your head office. Some companies have their own equipment and employ qualified fabric technicians to carry out the testing at their headquarters. When using an accredited testing house, it will not matter in which country the test is done, because the test has to be carried out to the agreed method, and no matter which country does the test, it will be done in exactly the same way. As the fabric construction is such an important part of the garment, when finalized, the fiber composition, weight, and yarn count should be added to your contract with the supplier.

The testing house will give you guidance on the result you should be achieving and the method that you should be using.

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A room has a width of 14.1 feet, a length of 15.5 feet, and a ceiling height of 12.0 ft. The average flow rate for this room's a

prohojiy [21]

Answer:

Your question lacks the time required hence i will calculate the Average flow rate using a general concept and an assumed time value of 25 seconds

ANSWER : 104.904 ft^3/sec

Explanation:

General concept : Average flow rate is the volume of fluid per unit time through an area

Hence the average flow rate of the air conditioning unit of this room

Volume of the room / time taken for the air to cycle the room = v / t

assuming the time taken = 25 seconds

volume of room = width * length * height

= 14.1 * 15.5 * 12 = 2622.6 ft^3

Average flow rate = V/ t

= 2622.6 / 25 = 104.904 ft^3/sec

8 0

3 years ago

Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K. The constant B = 3. 56 times 1014 9cm -3 K-3/2) and the band

meriva

This question is incomplete, the complete question is;

Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K.

The constant B = 3.56×10¹⁴ (cm⁻³ K^-3/2) and the bandgap voltage E = 1.42eV.

Answer: the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³

Explanation:

Given that;

T = 300k

B = 3.56×10¹⁴ (cm⁻³ K^-3/2)

Eg = 1.42 eV

we know that, the value of Boltzmann constant k = 8.617×10⁻⁵ eV/K

so to find the ni for gallium arsenide;

ni = B×T^(3/2) e^ ( -Eg/2kT)

we substitute

ni = (3.56×10¹⁴)(300^3/2) e^ ( -1.42 / (2× 8.617×10⁻⁵ 300))

ni = (3.56×10¹⁴)(5196.1524)e^-27.4651

ni = (3.56×10¹⁴)(5196.1524)(1.1805×10⁻¹²)

ni = 2.1837 × 10⁶ cm⁻³

Therefore the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³

4 0

3 years ago

In order to produce a certain semiconductor, a process called doping is performed in which phosphorus is diffused into germanium

Nimfa-mama [501]

Answer:

The diffusivity is given as $8.064\times 10^{-16} m^2/s$

Explanation:

From the given data as in the attached question found via the search (because the values were not clear in this one)

$D_o=2.0\times10^{-4}m^2/s\\Q_d=240.6 kJ/mol=2.406 \times 10^5 J/mol\\Gas Constant=R=8.314 J/mol K\\Temperature =830 C =830+273 =1103 K$

So the Diffusion coefficient is given as

$D=D_oe^{\frac{-Q_d}{RT}}\\D=(2.0\times 10^{-4})e^{\frac{-2.406\times 10^5}{8.314\times 1103}}\\D=8.0640\times 10^{-16} m^2/s$

So the diffusivity is given as $8.064\times 10^{-16} m^2/s$

4 0

4 years ago

In a constant-head permeability test in the laboratory, the length of the soil specimen is 275 mm and its cross sectional area i

sattari [20]

Answer:

The head difference across the soil specimen is 39.29 cm and the discharge velocity is 0.02 cm/s

Explanation:

The head difference across the soil specimen is:

$k=\frac{QL}{Aht}$

Where

k = hydraulic conductivity = 0.014 cm/s

Q = volume of water collected = 150 cm³/min = 2.5cm³/s

L = length of the soil specimen = 275 mm = 27.5 cm

A = area = 125 cm²

Replacing:

$0.014=\frac{2.5*27.5}{125*h*1} \\h=39.29cm$

The hydraulic gradient is:

$i=\frac{h}{L} =\frac{39.29}{27.5} =1.43$

The discharge velocity is:

v = k*i = 0.014 * 1.43 = 0.02 cm/s

8 0

3 years ago

Engineers please help im not good when it comes to drawing

kakasveta [241]

Answer:

I'm good at drawing and computer-animated design

Explanation:

3 0

3 years ago