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notka56 [123]
2 years ago
14

Two previously undeformed rod-shaped specimens of copper are to be plastically deformed by reducing their cross-sectional areas.

One has a circular cross section, and the other has a rectangular cross section. During deformation, the circular cross section is to remain circular, and the rectangular cross section is to remain rectangular. Their original and deformed dimensions are as follows:
Circular (diameter, mm) Rectangular(mm)
Original dimensions 18.0 20 x 50
Deformed dimension 15.9137 x55.1

Required:
Which of these specimens will be the hardest after plastic deformation, and why?
Engineering
1 answer:
mezya [45]2 years ago
4 0
I am not sure I am stuck on this and I have been for 45 min someone please help me and this girl or boy!!
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Which explanation best summarizes what went wrong during Paul’s cost analysis?
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1 year ago
A biotechnology company produced 225 doses of somatropin, including 11 which were defective. Quality control test 15 samples at
Radda [10]

Answer:

  • <u>0.59</u>

Explanation:

The <em>batch</em> is <em>rejected</em> if any of the <em>random samples are found defective</em>, or, what is the same, it will be accepted only if all 15 samples are good.

The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.

The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.

The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.

By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.

The joint probability of all the samples are good is the product of each probability:

\frac{214}{225}\cdot\frac{213}{224}\cdot\frac{212}{223}\cdot\frac{211}{222}\cdot\frac{210}{221}\cdot\frac{209}{220}\cdot\frac{208}{219}\cdot\frac{207}{218}\cdot\frac{206}{217}\cdot\frac{205}{216}\cdot\frac{204}{215}\cdot\frac{203}{214}\cdot\frac{202}{213}\cdot\frac{201}{212}\cdot\frac{200}{211}\cdot\frac{199}{210}

The result is: 0.41278 ≈ 0.41

The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.

Therefore, <em>the probability that the batch is rejected</em> is 1 - 0.41 = 0.59.

4 0
3 years ago
In the fully developed region of flow in a circular pipe, does the velocity profile change in the flow direction?
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<em>No, the velocity profile does not change in the flow direction.</em>

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In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, <em>then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.</em>

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If a shear stress acts in one plane of an element, there must be an equal and opposite shear stress acting on a plane that is
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90 degrees

Explanation:

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Therefore as per the given situation it should be 90 degrees from the plane

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