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aksik [14]
3 years ago
11

Describe the basic mechanism by which positive photolithography transfers a pattern, and comment on the chemical interactions of

the resist throughout the entire process.
Engineering
1 answer:
ankoles [38]3 years ago
6 0

Photo-lithography is the method of giving geometric shapes on a mask to the surface of a silicon wafer.

<u>Explanation:</u>

The fabrication of an integrated circuit (IC) requires a variety of physical and chemical processes conducted on a semiconductor (e.g., silicon) substrate. In common, the numerous methods used to create an IC fall into three divisions: film deposition, patterning, and semiconductor doping.

Films from both conductors (such as polysilicon, aluminum, and extended recently copper) and nonconductors (various forms of silicon dioxide, silicon nitride, and others) are utilized to combine and separate transistors and their parts.

Selective doping of different regions of silicon permits the conductivity of the silicon to be altered with the application of voltage. By building structures of these various parts millions of transistors can be assembled and wired together to form the complex circuitry of a modern microelectronic device.  

Fundamental to all of these methods is lithography, i.e., the development of three-dimensional relief images on the substrate for subsequent transfer of the model to the substrate.

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Definition of ceramics
dezoksy [38]

Answer:

Ceramics are generally made by taking mixtures of clay, earthen elements, powders, and water and shaping them into desired forms. Once the ceramic has been shaped, it is fired in a high temperature oven known as a kiln. Often, ceramics are covered in decorative, waterproof, paint-like substances known as glazes.

Explanation:

3 0
3 years ago
Read 2 more answers
A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

v₂² = 88.88

v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

hence , actual velocity will be equal to 8.39 m/s

6 0
3 years ago
A soil weighs 2,520 lbs/CY in its in situ condition, 1,970 lb/CY in its loose condition after excavation, and 3,025 lbs/CY in an
azamat

Answer:

load factor = 0.782

Shrink Factor = 0.833

no of truck is  62500

Explanation:

given data

soil weighs in situ condition  =  2,520 lbs/CY

soil weighs in loose condition  =  1,970 lb/CY

soil weighs in embanked state = 3,025 lbs/CY

average volume = 16 LCY

soil from a borrow pit = 1 million CCY

solution

first we get here Load Factor that is express as

load factor = \frac{1,970}{2550}

load factor = 0.782

and Shrink Factor will be as

Shrink Factor = \frac{2520}{3025}

Shrink Factor = 0.833

and

no of truck will be

no of truck = \frac{1000000}{16}

no of truck is  62500

6 0
4 years ago
A lake is fed by a polluted stream and a sewage outfall. The stream and sewage wastes have a decay rate coefficient (k) of 0.5/d
joja [24]

Solution :

Given :

k = 0.5 per day

$C_s = 10 \ mg/L \ ; \ \ Q_s= 40 \ m^3/s$

$C_{sw} = 100 \ ppm \ ; \ \ Q_{sw}= 0.5 \ m^3/s$

Volume, V $= 200 \ m^3$

Now, input rate = output rate + KCV ------------- (1)

Input rate  $= Q_s C_s + Q_{sw}C_{sw}$

                $=(40 \times 10) + (0.5\times 100)$

                $= 2 \times 10^5 \ mg/s$

The output rate $= Q_m C_{m}$

                          = ( 40 + 0.5 ) x C x 1000

                          $=40.5 \times 10^3 \ C \ mg/s$

Decay rate = KCV

∴$KCV =\frac{0.5/d \times C \  \times 200 \times 1000}{24 \times 3600}$

            = 1.16 C mg/s

Substituting all values in (1)

$2 \times 10^5 = 40.5 \times 10^3 \ C+ 1.16 C$

C = 4.93 mg/L

4 0
3 years ago
Explain why the following scenario fails to meet the definition of a project description.
s344n2d4d5 [400]

Answer:

The youth hockey training facility

Explanation:

7 0
3 years ago
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