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MA_775_DIABLO [31]

3 years ago

10

The theoretical maximum specific gravity of a mix at 5.0% binder content is 2.495. Using a binder specific gravity of 1.0, find

the theoretical maximum specific gravity at 6.5% binder content. (Hint: Both use the same stone, therefore have the same Gmb)

1 answer:

PSYCHO15rus [73]3 years ago

3 0

Answer:

The theoretical maximum specific gravity at 6.5% binder content is 2.44.

Explanation:

Given the specific gravity at 5.0 % binder content 2.495

Therefore

95 % mix + 5 % binder gives S.G. = 2.495

Where the binder is S.G. = 1, Therefore

Per 100 mass unit we have (Mx + 5)/(Vx + 5) = 2.495

(95 +5)/(Vx +5) = 2.495

2.495 × (Vx + 5) = 100

Vx =35.08 to 95

Or density of mix = Mx/Vx = 95/35.08 = 2.7081

Therefore when we have 6.5 % binder content, we get

Per 100 mass unit

93.5 Mass unit of Mx has a volume of

Mass/Density = 93.5/2.7081 = 34.526 volume units

Therefore we have

At 6.5 % binder content.

(100 mass unit)/(34.526 + 6.5) = 2.44

The theoretical maximum specific gravity at 6.5% binder content = 2.44.

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The dry unit weight of a soil sample is 14.8 kN/m3.

Jlenok [28]

Answer:

See attachment for completed question

Explanation:

Given that; Brainly.com

What is your question?

mkasblog

College Engineering 5+3 pts

The dry unit weight of a soil sample is 14.8 kN/m3.

Given that G_s = 2.72 and w = 17%, determine:

(a) Void ratio

(b) Moist unit weight

(c) Degree of saturation

(d) Unit weight when the sample is fully saturated

See complete solving at attachment

4 0

3 years ago

A cast-iron tube is used to support a compressive load. Knowing that E 5 10 3 106 psi and that the maximum allowable change in l

Papessa [141]

Answer:

(a) 2.5 ksi

(b) 0.1075 in

Explanation:

(a)

$E=\frac {\sigma}{\epsilon}$

Making $\sigma$ the subject then

$\sigma=E\epsilon$

where $\sigma$ is the stress and $\epsilon$ is the strain

Since strain is given as 0.025% of the length then strain is $\frac {0.025}{100}=0.00025$

Now substituting E for $10\times 10^{6} psi$ then

$\sigma=(10\times 10^{6} psi)\times 0.00025=2500 si= 2.5 ksi$

(b)

Stress, $\sigma= \frac {F}{A}$ making A the subject then

$A=\frac {F}{\sigma}$

$A=\frac {\pi(d_o^{2}-d_i^{2})}{4}$

where d is the diameter and subscripts o and i denote outer and inner respectively.

We know that $2t=d_o - d_i$ where t is thickness

Now substituting

$\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}$

$\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4$

$(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

But the outer diameter is given as 2 in hence

$(2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

$2^{2}-(\frac {1600}{2500\times \pi}\times 4)=d_i^{2}$

$d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in$

As already mentioned, $2t=d_o - d_i hence t=0.5(d_o - d_i)$

$t=0.5(2-1.785)=0.1075 in$

3 0

3 years ago

4. Two technicians are discussing the evaporative emission monitor. Technician A says that serious monitor faults cause a blinki

snow_lady [41]

Answer:

The correct option is;

Neither Technician A nor B

Explanation:

The evaporative emission monitor or Evaporaive Emission Control System EVAP System monitors enables the Power Control Module of the car to check fuel system leak integrity and the vapor consumption efficiency during engine combustion

It is a requirement of EPA on cars to check the emission of smug forming evaporates from cars

Serious monitor faults can cause the turning on of the check engine lights and the vehicle will not pass OBD II test, but it will not lead to engine shutdown

It runs when the engine is 15 to 85% full and the TP sensor is between 9% and 35%.

Therefore, the correct option is that neither Technician A nor B are correct.

3 0

3 years ago

2.4: Add a method called setValue(), and the description of setValue is: public int setValue(long searchKey) In this method, the

Yanka [14]

Answer:

Below is java code that must be used for the given question:

// highArray.java

// demonstrates array class with high-level interface

// to run this program: C>java HighArrayApp

////////////////////////////////////////////////////////////////

class HighArray

{

private long[] a; // ref to array a

private int nElems; // number of data items

//-----------------------------------------------------------

public HighArray(int max) // constructor

{

a = new long[max]; // create the array

nElems = 0; // no items yet

}

//-----------------------------------------------------------

public setValue find(long searchKey)

{ // find specified value

int j;

for(j=0; j<nElems; j++) // for each element,

if(a[j] == searchKey) // found item?

break; // exit loop before end

if(j == nElems) // gone to end?

return false; // yes, can't find it

else

return true; // no, found it

} // end find()

//-----------------------------------------------------------

public void insert(long value) // put element into array

{

a[nElems] = value; // insert it

nElems++; // increment size

}

//-----------------------------------------------------------

public void display() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element,

System.out.print(a[j] + " "); // display it

System.out.println("");

}

//-----------------------------------------------------------

} // end class HighArray

////////////////////////////////////////////////////////////////

class HighArrayApp

{

public static void main(String[] args)

{

int maxSize = 100; // array size

HighArray arr; // reference to array

arr = new HighArray(maxSize); // create the array

arr.insert(77); // insert 10 items

arr.insert(99);

arr.insert(44);

arr.insert(55);

arr.insert(22);

arr.insert(88);

arr.insert(11);

arr.insert(00);

arr.insert(66);

arr.insert(33);

arr.display(); // display items

int searchKey = 35; // search for item

if( arr.find(searchKey) )

System.out.println("Found " + searchKey);

else

System.out.println("Can't find " + searchKey);

} // end main()

} // end class HighArrayApp

Explanation:

6 0

3 years ago

A silicon diode has a saturation current of 6 nA at 25 degrees Celcius. What is the saturation current at 100 degrees Celsius?

Illusion [34]

Answer:

0.0659 A

Explanation:

Given that :

$I_{0} = 6nA$ ( saturation current )

at 25°c = 300 k ( room temperature )

n = 2 for silicon diode

Determine the saturation current at 100 degrees = 373 k

Diode equation at room temperature = I = Io $\frac{V}{e^{0.025*n} }$

next we have to determine the value of V at 373 k

q / kT = (1.6 * 10^-19) / (1.38 * 10^-23 * 373) = 31.08 V^-1

Given that I is constant

Io = $\frac{e^{0.025*2} }{31.08}$ = 0.0659 A

3 0

3 years ago