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Zinaida [17]
3 years ago
6

I really need help on this!

Engineering
2 answers:
Dafna1 [17]3 years ago
8 0
C: benchmark because I have done this before
den301095 [7]3 years ago
3 0
The answer is D!
Plz mark brainliest
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How can statistical analysis of a dataset inform a design process
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Answer:

Explanation:.

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Are front-end engineers starting to decline in China?
laiz [17]

Yes. They are declining in China. Very fast

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The proposed grading at a project site will consist of 25,100 m3 of cut and 23,300 m3 of fill and will be a balanced earthwork j
Anna [14]

Answer:

the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL

Explanation:

Given that;

volume of cut = 25,100 m³

Volume of dry soil fill = 23,300 m³

Weight of the soil will be;

⇒ 93% × 18.3 kN/m³ × 23,300 m³

= 0.93 × 426390 kN 3

= 396,542.7 kN  

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Required amount of moisture = (12.9 - 8.3)% = 4.6 %

So,

Weight of water required = 4.6% × 396,542.7 = 18241 kN

Volume of water required = 18241 / 9.81 = 1859 m³

Volume of water required = 1859 kL

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6 0
3 years ago
What is a Flame Front Generator?
Inessa [10]

Answer and Explanation:

Flame Front Generator: It is a ignition system which is very useful in flaring system .In this system the air and gases are mixed together and make a combustible air gas mixture. There is a flame front region where the combustion reaction takes place , it is the region where gases as like hydrogen and air mixed with each other and form combustible gases.

7 0
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For a turning operation, you have selected a high-speed steel (HSS) tool and turning a hot rolled free machining steel. Your dep
Alisiya [41]

Answer:

MRR = 1.984

Explanation:

Given that                              

Depth of cut ,d=0.105 in

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feed f= 0.015 ipr

Now  the metal   removal  rate   given as

MRR= 12 f V d

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V= Speed

f=Feed

MRR= Metal removal rate

By putting the values

MRR= 12 f V d

MRR = 12 x 0.015 x 105 x 0.105

MRR = 1.984

Therefore answer is -

1.944

8 0
3 years ago
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