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lesya692 [45]
3 years ago
14

A. Briefly describe the microstructural difference between spheroidite and tempered martensite. Explain why tempered martensite

is much stronger.
b. Explain why tempered martensite is much harder and stronger.
Engineering
1 answer:
masha68 [24]3 years ago
4 0

Answer:

Answered below.

Explanation:

A) Both spheroidite & tempered martensite possess sphere - like cementite particles within their microstructure known as a ferrite matrix. However, the difference is that these particles are much larger for spheroidite than tempered.

B) Tempered martensite is much harder and stronger than spheroidite primarily because there is much more ferrite - cementite phase boundary area for its sphere - like cementite particles.

This is because the greater the boundary area, the more the hardness.

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Base course aggregate has a target dry density of 119.7 Ib/cu ft in place. It will be laid down and compacted in a rectangular s
djyliett [7]

Answer:

total weight of aggregate =  5627528 lbs = 2814 tons  

Explanation:

we get  here volume of space to be filled with aggregate that is

volume = 2000 × 48 × 0.5

volume = 48000 ft³

now filling space with aggregate of the density that is

density = 0.95 × 119.7

density = 113.72 lb/ft³

and dry weight of this aggregate is

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dry weight = 5458320 lbs

we consider here percent moisture is by weigh

so weight of moisture in aggregate will be

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weight of moisture = 169208 lbs

so here total weight of aggregate is

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3 years ago
What is the difference between a refrigeration cycle and a heat pump cycle?
sukhopar [10]

Answer:

In refrigeration cycle heat transfer from inside refrigeration

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Explanation:

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3 years ago
Give an example of one technology that is well matched to the needs of the environment, and one technology that is not.
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An automated transfer line is to be designed. Based on previous experience, the average downtime per occurrence = 5.0 min, and t
IRINA_888 [86]

Answer:

a) 28 stations

b) Rp = 21.43

E = 0.5

Explanation:

Given:

Average downtime per occurrence = 5.0 min

Probability that leads to downtime, d= 0.01

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a) For the optimum number of stations on the line that will maximize production rate.

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Tp = Tc + Ftd

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At minimum pt. = 0, we have:

dTp/dn = 0

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Solving for n²:

n^2 = \frac{39.2}{0.05} = 784

n = \sqrt{784} = 28

The optimum number of stations on the line that will maximize production rate is 28 stations.

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Answer:

true

Explanation:

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