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AnnyKZ [126]
3 years ago
15

A 2m high sandy fill material was placed loosely at a relative density of 47%. Laboratory studies indicated that the maximum and

minimum void ratios of the fill material are 0.92 and 0.53, respectively. Construction specifications required that the fill be compacted to a relative density of 80%.
If Gs = 2.65 determine:

a) the dry unit weight of the fill before and after compaction
b) the final height of the fill after compaction
Engineering
1 answer:
Llana [10]3 years ago
6 0

Answer:

2

Explanation:To find the solution we must first make the transformations to international units,

In this way:

Inches of length (L) to feet =

So we can identify the area by hour:

Once the area is identified, we now identify the number of passes required,

Passes required = Area x number of passes

We transform speed to international units

So we can identify the covered area

where

m = drum widht

p = efficiency

In this way we obtain the number of rollers given by:

No. of Rollers = Passes required / covered area

No of rollers = 48136/40269

No of rollers = 1.19, that is, 2

Read more on Brainly.com - brainly.com/question/13686900#readmore.

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A MOSFET differs from a JFET mainly because
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Answer:

The answer is option

C . the JFET has a PN junction

Explanation:

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Which Finance jobs can someone pursue with only a high school diploma? Check all that apply.
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What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?
Yakvenalex [24]

Answer:

Activation\ Energy=2.5\times 10^{-19}\ J

Explanation:

Using the expression shown below as:

N_v=N\times e^{-\frac {Q_v}{k\times T}

Where,

N_v is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = 1.38\times 10^{-23}\ J/K

{Q_v} is the activation energy

T is the temperature

Given that:

N_v=2.3\times 10^{13}

N = 10 moles

1 mole = 6.023\times 10^{23}

So,

N = 10\times 6.023\times 10^{23}=6.023\times 10^{24}

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (425 + 273.15) K = 698.15 K  

T = 698.15 K

Applying the values as:

2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

Q_v=2.5\times 10^{-19}\ J

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3 years ago
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