What are materials engineers trying to discover when they study different materials? Whether or not materials corrode and oxidize how materials perform and deteriorate if certain materials have been used in the past and at which historical ages which metals conduct heat

Answers

Answer 1

Answer:

Material engineers study various materials to discover the reason and cause of its existence, the chemical properties, how long it has been there, and how it impacts human life.

Explanation:

Material engineer is an engineering discipline that focuses to improve human life by studying the environment and the various elements or materials it holds.

It uses the power of pure science to test and analysis its findings, documents its features and exposes them to the world. Materials like metals and other elemental forms were all tested by these engineers to determine its history and chemical and biological use.

Answer 2

Answer:

how materials perform and deteriorate

Explanation:


Related Questions

What are the common approximations made in the analysis of heat exchangers?

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Answer: making sure that they are up to date

Explanation:

What is a specialized accreditation? A. evaluation of the quality of instruction B. evaluation of a particular program C. evaluation of students studying in an organization D. evaluation of recreational facilities in an organization

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Answer:

B. evaluation of a particular program

Explanation:

Before students enrol into any given discipline, they should first ensure that their program of choice is well accredited. In a specific program, specialized accreditation has the function of telling would be students if the program meets up with academic standards in the field

Accreditation is a way of assessing faculty and curriculum quality of schools to make sure that they are up to academic standards and are also preparing students to future success in the field.

What are the main causes of injuries when using forklifts?

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The forklift overturning is a very common way of getting injured from a forklift. Overturning the forklift means it tips over onto it's side due to the operator turning it too fast.

how do you make coke for steel?

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Can you be a bit more specific plz and that will let me identify the answer

The production process of rods from machine "A" yields specimen with the following specs. Mean: µ(LA)=20.00mm, STD: s(LA)=0.50mm. You need to purchase new machine to increase the production capacity and accuracy together. If the new machine "B" will produce half of the total rods, what is the STD, s(LB), that needs to achieve the total STD, s(LT)=0.4mm? Assume Corr(A,B)=0.4

Answers

Answer: the standard deviation STD of machine B is s (Lb) = 0.4557

Explanation:

from the given data, machine A and machine B produce half of the rods

Lt = 0.5La + 0.5Lb

so

s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)

but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)

so we substitute

s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)

0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)

0.64 = 0.25 + s²(Lb) + 0.4s(Lb)

s²(Lb) + 0.4s(Lb) - 0.39 = 0

s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2

s (Lb) = 0.4557

therefore the standard deviation STD of machine B is s (Lb) = 0.4557

A tensile test uses a test specimen that has a gage length of 50 mm and an area = 206 mm2. During the test, the specimen yields under a load of 97,944 N. The corresponding gage length = 50.2 mm - this is at the 0.2 percent yield point. The maximum load of 162,699 N is reached at a gage length = 63 mm. If fracture occurs at a gage length of 71 mm, determine the percent elongation in % - enter your answer as a whole number, not as a fraction.

Answers

Answer:

The percent elongation in the length of the specimen is 42%

Explanation:

Given that:

The gage length of the original test specimen  [tex]L_o[/tex] = 50 mm

The final gage length [tex]L_f[/tex] = 71 mm

The area = 206 mm²

maximum load  =  162,699 N

To determine the percent elongation in %, we use the formula:

[tex]\%EL = \dfrac{L_f-L_o}{L_o}\times 100[/tex]

[tex]\%EL = \dfrac{71 \ mm-50 \ mm}{50 \ mm}\times 100[/tex]

[tex]\%EL = \dfrac{21 mm}{50 \ mm}\times 100[/tex]

[tex]\%EL = 0.42 \times 100[/tex]

[tex]\mathbf{\%EL = 42 \%}[/tex]

The percent elongation in the length of the specimen is 42%

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