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kicyunya [14]
1 year ago
8

Is annealed polycrystalline copper a material that work hardens significantly or a material that exhibits a low work-hardening r

ate
Engineering
1 answer:
Olegator [25]1 year ago
6 0

Those that harden under strain, such as the aluminum-magnesium alloys used in beverage cans and the copper-zinc alloy, brass, used for cartridges, which show more strain hardening than pure copper or aluminum, respectively.

When a material is deformed under a substantial amount of strain, strain hardening is seen as a strengthening process. Lamellar crystals and chain molecule orientation on a vast scale are the culprits. When plastic materials are stretched past their yield point, this phenomena is frequently seen. When a metal is stretched past its yield point, strain hardening occurs. The metal appears to get stronger and harder to deform as more stress is needed to cause additional plastic deformation. Strain hardening is directly related to fatigue.

Learn more about strain hardening here-

brainly.com/question/15058191

#SPJ4

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Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to b
coldgirl [10]

Answer:

A) q'_free = 3146.41 W/m²

B) q'_forced = 7521.41 W/m²

Explanation:

We are given;

Free convection coefficient; h_fr = 5 W/m²K

Force convection coefficient; h_forced = 30 W/m²K

Emissivity; ε = 0.95

Temperature of surrounding which is equal to temperature of air; T_s = T_air = 200°C = 473K

Initial temperature; T_i = 25°C = 298K

A) Now, since the convection feature is disabled, the mode of heat transfer associated with this condition is through free convection and radiation.

Thus, the formula for the heat flux under this condition is given as;

q'_free = q'_free convection + q'_radiation

q'_free convection = h_free(T_∞ - T_i) where T_∞ is equivalent to the value of T_air

Also, q'_radiation = ε•σ((T_air)⁴ - (T_i)⁴)

Where, σ is stephan boltzmann constant and has a constant value of 5.67 × 10^(−8) W/m²K⁴

Thus, rewriting;

q'_free = q'_free convection + q'_radiation

We have;

q'_free = [h_free(T_∞ - T_i)] + [ε•σ((T_air)⁴ - (T_i)⁴)]

Plugging in the relevant values to obtain;

q'_free = [5(473 - 298)] + [0.95•5.67 × 10^(−8)((473)⁴ - (298)⁴)]

q'_free = 875 + 2271.41

q'_free = 3146.41 W/m²

B) Now, in this case, since the convection feature is disabled, the mode of heat transfer associated with this condition is through forced convection and radiation.

Thus, the formula for the heat flux under this condition is given as;

q'_forced = q'_forced convection + q'_radiation

Where;

q'_forced convection = h_forced(T_∞ - T_i) where T_∞ is equivalent to the value of T_air

Also, q'_radiation = ε•σ((T_air)⁴ - (T_i)⁴)

Thus, rewriting;

q'_forced = q'_free convection + q'_radiation

We have;

q'_forced = [h_forced(T_∞ - T_i)] + [ε•σ((T_air)⁴ - (T_i)⁴)]

Plugging in the relevant values to obtain;

q'_forced = [30(473 - 298)] + [0.95•5.67 × 10^(−8)((473)⁴ - (298)⁴)]

q'_forced = 5250 + 2271.41

q'_forced = 7521.41 W/m²

7 0
4 years ago
Interpret the Blame responsibility and causation in your own words in the light of Columbia Accident.
Licemer1 [7]

Answer:

Proposed Improvements and Generic Lessons

Within 2 h of losing the signal from the returning spacecraft, NASA’s Administrator established the Columbia Accident Investigation Board (CAIB) to uncover the conditions that had produced the disaster and to draw inferences that would help the US space program to emerge stronger than before (CAIB, 2003). Seven months later, the CAIB released a detailed report that included its recommendations (Starbuck and Farjoun, 2005).

The CAIB (2003) report attempted to seek answers to the following four crucial questions:

1.

Why did NASA continue to launch spacecraft despite many years of known foam debris problems?

2.

Why did NASA managers conclude, despite the concerns of their engineers, that the foam debris strike was not a threat to the safety of the mission?

3.

How could NASA have forgotten the lessons of Challenger?

4.

What should NASA do to minimize the likelihood of such accidents in the future?

Although the CAIB’s comprehensive report raised important questions and offered answers to some of them, it also left many major questions unanswered (Starbuck and Farjoun, 2005).

1.

Why did NASA consistently ignore the recommendations of several review committees that called for changes in safety organization and practices?

2.

Did managerial actions and reorganization efforts that took place after the Challenger disaster contribute, both directly and indirectly, to the Columbia disaster?

3.

Why did NASA’s leadership fail to secure more stable funding and to shield NASA’s operations from external pressures?

By examining, with respect to the Columbia disaster, the case of NASA as an organization, one can try to extract generalizations that could be useful for other organizations, especially those engaged in high-risk activities—such as nuclear power plants, oil and gas, hospitals, airlines, armies, and pharmaceutical companies—and such generic principles may also be salutary for any kind of organization.

The CAIB (2003) report recommended developing a plan to inspect the condition of all RCC systems, the investigation having found the existing inspection techniques to be inadequate. RCC panels are installed on parts of the shuttle, including the wing leading edges and nose cap, to protect against the excessive temperatures of reentry. They also recommended that taking images of each shuttle while in orbit should be standard procedure as well as upgrading the imaging system to provide three angles of view of the shuttle, from liftoff to at least SRB separation. “The existing camera sites suffer from a variety of readiness, obsolescence, and urban encroachment problems.” The board offered this suggestion because NASA had had no images of the Columbia shuttle clear enough to determine the extent of the damage to the wing. They also recommended conducting inspections of the TPS, including tiles and RCC panels, and developing action plans for repairing the system. The report included 29 recommendations, 15 of which the board specified must be completed before the shuttle returned to flight status, and also made 27 “observations” (CAIB, 2005).

7 0
3 years ago
How deep does electrical conduit need to be buried?
GenaCL600 [577]
In general, bury metal conduits at least 6 inches below the soil surface. You may also run them at a depth of 4 inches under a 4-inch concrete slab. Under your driveway, the conduits must be below a depth of 18 inches, and under a public road or alleyway, they must be buried below 24 inches.
7 0
3 years ago
The pressure at the bottom of an 18 ft deep storage tank for gasoline is how much greater than at the top? Express your answer i
Julli [10]
5.85 psig

Using a specific gravity of 0.75 as an average for red\automobile gasoline.

Water at standard conditions (60 degF) is 2.31 feet = 1 psig

80/2.31 then multiply x .75 to compensate for specific gravity of water being 1.0
4 0
3 years ago
Electric current originates from which part of an atom? *
yanalaym [24]

Answer: Electric current originates from positively charged protons negatively charged electrons of an atom.

Explanation:

The movement of ions (positive or negative) from one point to another is called electric current.

An atom has three sub-atomic particles. These are protons, neutrons and electrons.

Protons are positively charged, neutrons have no charge and electrons are negatively charged. Protons and neutrons reside inside the nucleus of an atom whereas electrons revolve around the nucleus.

So, protons and electrons are responsible for originating electric current form an atom as these are the charged particles.

Thus, we can conclude that electric current originates from positively charged protons negatively charged electrons of an atom.

3 0
3 years ago
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