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vladimir1956 [14]
3 years ago
6

What are some constraints related to size, time, or materials? Remember, constraints are limitations, issues, or obstacles.

Engineering
1 answer:
Sergeeva-Olga [200]3 years ago
4 0

Answer:

too big, too small, deadlines, not enough materials

Explanation:

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If a material is found to be in the tertiary phase of creep, the following procedure should be implemented:
kirill [66]

When a material is found to be in the tertiary phase of creep, the following procedure should be implemented that is the component should be replaced immediately. Therefore, Option C is correct.

<h3>What do you mean by a tertiary degree of creep?</h3>

Tertiary Creep has an extended creep rate and terminates when the material breaks or ruptures. It is related to each necking and formation of grain boundary voids. The wide variety of possible stress-temperature- time combos is infinite.

Therefore, When a material is found to be in the tertiary phase of creep, the following procedure should be implemented that is the component should be replaced immediately. Option C is correct.

Learn more about creep:

brainly.com/question/10565749

#SPJ1

8 0
2 years ago
Jason’s Petrochems is a firm that brings in chemical intermediates such as organic fertilizers and organic solvents from differe
saul85 [17]

Answer:

an importer

Explanation:

Jason's petrochems depends on imported intermediates for its production/processing. From the question, the firm imports chemical intermediates, processes them in its laboratories before selling them. It also uses the intermediates to produce various fertilizers and solvents. Essentially its feed stocks ( inputs) are wholly imported from different countries. Hence the firm is an import dependent petrochemical complex.

6 0
3 years ago
During the reaction, 3.50 μmol of HCl are produced. Calculate the final pH of the reaction solution. Assume that the HCl is comp
lyudmila [28]

Answer:

The pH of the solution will be equal to 5.46

Explanation:

The dissociation reaction of HCl is equal to:

HCl → H+ + Cl-

To solve the exercise we must first convert the µmoles to moles using the following conversion factor:

3.5µmoles x \frac{1 mol}{1x10^{6} umol} = 3.5x10^{-6}moles

Assuming a liter of solution, we can calculate the molar concentration by:

M = \frac{Number of moles}{Liter of solution}

Replacing:

M = \frac{3.5x10^{-6}moles }{1 L} = 3.5x10^{-6}moles/L

As this acid dissociates completely, the concentration of protons and chloride will be equal to 3.5x10^{-6}moles/L

The pH will be equal to:

pH = -log[H+]

Replacing:

pH = -log[3.5x10^{-6}] = 5.46

8 0
4 years ago
A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a
kirill [66]

Answer:

105.70 mm

Explanation:

Poisson’s ratio, v is the ratio of lateral strain to axial strain.

E=2G(1+v) where E is Young’s modulus, v is poisson’s ratio and G is shear modulus

Since G is given as 25.4GPa, E is 65.5GPa, we substitute into our equation to obtain poisson’s ratio

\begin{array}{l}\\65.5{\rm{ GPa}} = 2\left( {25.4{\rm{ GPa}}} \right)(1 + \upsilon )\\\\\upsilon = 0.2893\\\end{array}

Original length L_(i}

\upsilon = - \left( {\frac{{\left( {\frac{{{d_f} - {d_i}}}{{{d_i}}}} \right)}}{{\left( {\frac{{{L_f} - {L_i}}}{{{L_i}}}} \right)}}} \right)

Where d_{f} is final diameter, d_{i} is original diameter, L_{f} is final length and L_{i} is original length.

\begin{array}{l}\\0.2893 = - \left( {\frac{{\left( {\frac{{30.04{\rm{ mm}} - {\rm{30 mm}}}}{{{\rm{30 mm}}}}} \right)}}{{\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right)}}} \right)\\\\\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right) = - 4.6088 \times {10^{ - 3}}\\\end{array}

\begin{array}{l}\\105.2 - {L_i} = - \left( {4.6088 \times {{10}^{ - 3}}} \right){L_i}\\\\105.2 = 0.9953{L_i}\\\\{L_i} = 105.70{\rm{ mm}}\\\end{array}

Therefore, the original length is 105.70 mm

7 0
4 years ago
In a paragraph of 125 words, distinguish the engineering design process from the reverse engineering process, citing at least th
Yuki888 [10]

The difference between the engineering design process and the reverse engineering process are there, but very subtle. The engineering design process is a process based on careful planning and months of designing to create a blueprint for a certain project. The reverse engineering process is the base of learning to create something by working backwards on a previously made inspiration for your project. Three differences are that that the engineering design process doesn't require any physical learning and is more based on mental and written learning (unlike the reverse engineering process). Another difference includes that in the reverse engineering process you don't need to take anything apart and learn how things work based on the on hands aspect of this variety of engineering. The final difference is that the engineering design project is made fresh from your own thoughts and not based off of a similar project.

7 0
3 years ago
Read 2 more answers
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