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grin007 [14]
3 years ago
5

What does a graphic designer need to do to monitor the progress of the project​

Engineering
1 answer:
Sholpan [36]3 years ago
5 0

Answer:

At first the graphic designer has to break the process down into it's fundamental components. Which could be:

1) Discussion and Clear Comprehension of the Brief

2) Research and Collection of Information

3) Sketching

4) Execution and Experimentation

5) Finalization

6) Revision and Further Corrections

7) Submissions

Then self-analysis is required to figure out the current stage process. Project management soft-wares such as Trello could be helpful when it comes to tracking the progress. Hope this answers the question.

Explanation:

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Technician A says a solenoid is not used on modern automobiles. Technician B says fuel injectors and starter motor solenoids are
julia-pushkina [17]

Answer:

ON THE BOTTOM

Explanation:

red cord plus black cord

5 0
3 years ago
PLZ ASAP WILL GIVE BRAINLIST
gavmur [86]

Answer: A, B, C & F (interacting w computers, making decisions & solving problems, evaluating information & getting information).

Explanation: Those are the correct & verified answers.

7 0
3 years ago
Read 2 more answers
Question 7.1: Two possible overhead valve combustion chambers are being considered – the first has two valves; the second has fo
AleksandrR [38]

Answer:

1) The adoption of the second design we can see that the total valve perimeter is increased by 60.8%

2) Increase in flow are : 29%

3) Additional benefits in using 4 valves per cylinder:

a)For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

Explanation:

1) Perimeter of the first possible overhead valve combustion chamber with two valves:

P₂ = πd = π × 23 = 72.26mm

Perimeter of the second possible overhead valve combustion chamber with four valves:

P₄ = π2d = π × 18.5 × 2 = 116.24 mm

If second design is adopted, percentage increase = ((P₄ - P₂)/P₂)×100

     = ((116.24 - 72.26)/72.26)×100 = 0.6086 ×100 = 60.86%

Therefore, the total valve perimeter is shown to have increased by 60.8%

2) Formula for flow Area (A) = P × L = πkd²

Area of the first possible overhead valve combustion chamber with two valves: A₂ = πkd² = πk(23)² = 1662k mm²

Area of the first possible overhead valve combustion chamber with four valves: A₄ = πkd² = 2πk(18.5)² = 2150k mm²

The percentage increase in flow area: ((A₄ - A₂)/A₄)×100 = ((2150 - 1662)/2150)×100 = 29%

3) The additional benefits of using are:

a) For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

           

7 0
3 years ago
¿Qué aditivo se debe incorporar a la masa de hormigón para aumentar su resistencia frente a los ciclos alternados de hielo-deshi
tamaranim1 [39]

Answer:

Los aditivos que deben incorporarse a la masa de concreto para aumentar su resistencia a los ciclos alternos de congelación y descongelación son;

1. Agentes de arrastre de aire (AEA) o

2. Materiales poliméricos súper absorbentes

Explanation:

La resistencia alterna de los ciclos de congelación y descongelación en el concreto puede aumentarse mediante la adición de agentes de arrastre de aire.(AEA) que es un surfactante, crea burbujas de aire muy pequeñas en el concreto resultante para mejorar la durabilidad y resistencia del cemento al ciclo repetido de congelación y descongelación o materiales poliméricos súper absorbentes

Ejemplos de agentes de arrastre de aire son;

Sulfonatos alcalinos

Acidos de resinas sulfonadas

Sales de ácidos grasos

Ejemplos de materiales poliméricos superabsorbentes son;

SAP0.26CT

SAP0.39PT.

6 0
4 years ago
1. The equilibrium number of vacancies in Ni at 1123 K is 4.7x1022m-3. The atomic weight and density of Ni at 1123 K are 58.69 g
Ahat [919]

Answer:

vacancy formation energy of Ni is 1.400 eV

Explanation:

given data

number of vacancies in Ni = 4.7 x 10^{22}  m^{-3}

atomic weight = 58.69 g/mol

density = 8.8 g/cm³  

solution

we get here N that is

N  = \frac{N_A \times \rho}{A}   ...........1

N = \frac{6.023\times 10^{23} \times 8.8 \times 10^6}{58.69}

N = 9.030 \times 10^{28}  

and here no of vacancy will be

Nv = N \times e^{\frac{-Qv}{kT}}  .................2

put here value

4.7 \times 10^{22} = 9.030 \times 10^28 \times e^{\frac{-Qv}{8.62\times 10^{-5}\times 1123}}  

10^{-7} \times 5.20487 = e^{\frac{-Qv}{0.0968}}

take ln both side

ln (10^{-7} \times 5.20487 ) = ln (e^{\frac{-Qv}{0.0968}})

-14.468 = \frac{-Qv}{0.0968}  

Qv = 1.400 eV

so vacancy formation energy of Ni is 1.400 eV

3 0
3 years ago
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