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kherson [118]
3 years ago
5

Answer the following questions, and very briefly explain your answer:

Engineering
1 answer:
alexandr1967 [171]3 years ago
5 0

Answer:

A) micro defects are left behind on the surface of metal components during the manufacturing process. These defects, in the form of micro-cracks or pits, becomes initiation sites for crack propagation or corrosion. Removing these imperfections on the surface of metal parts by electroplating greatly improves the life of metal components.

B) it will reduce fatigue crack growth.

Dispersion hardening involves the inclusion of small, hard particles in the metal, thus restricting the movement of dislocations, and thereby raising the strength properties. In dispersion hardening it is assumed that the precipitates do not deform with the matrix and that a moving dislocation bypasses the obstacles (precipitates) by moving in the clean pieces of crystal between the precipitated particles.

C) stress concentrations such as changes in section with sharp corners caused yielding, which will typically occur first at a stress concentration. For ductile materials localised plastic deformation can cause a redistribution of stress, enabling the component to continue to carry load. Brittle materials will typically fail at the stress concentration. Repeated loading may cause a fatigue crack to initiate and slowly grow at a stress concentration leading to the failure of even ductile materials. Fatigue cracks always start at stress raisers, so removing such defects increases the fatigue strength.

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Which of the following is NOT an ASE certification? Select one:
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The term ASE is known to be a body that tends to promotes excellence in regards to vehicle repair, service as well as parts distribution.

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2 years ago
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svetoff [14.1K]

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3 years ago
Please answer the questions !
gizmo_the_mogwai [7]

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Explanation:

6 0
3 years ago
Showing all of your work and algebra,generate an approximate expression for T as a function ofthe other variables. (b) Explain w
shusha [124]

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5 0
3 years ago
Consider a resistor made of pure silicon with a cross-sectional area pf 0.5 μm2, and a length of 50 μm. What is the resistance o
lukranit [14]

Answer: 24 pA

Explanation:

As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.

Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵  Ω  cm.

The resistance R of a given resistor, is expressed by the following formula:

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7 0
3 years ago
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