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Soloha48 [4]
3 years ago
15

An incompressible fluid flows along a 0.20-m-diameter pipe with a uniform velocity of 3 m/s. If the pressure drop between the up

stream and downstream sections of the pipe is 20 kPa, determine the power transferred to the fluid due to fluid normal stresses.
Engineering
1 answer:
KIM [24]3 years ago
6 0

Answer:

The power transferred to the fluid is 1.8852 kW

Explanation:

Power = pressure drop × area × velocity

pressure drop = 20 kPa

area = πd^2/4 = 3.142×0.2^2/4 = 0.03142 m^2

velocity = 3 m/s

Power = 20 × 0.03142 × 3 = 1.8852 kJ/s = 1.8852 kW

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How can I find the quotient of 27 divided 91.8
gayaneshka [121]
Step 1:
Start by setting it up with the divisor 8 on the left side and the dividend 27 on the right side like this:

8 ⟌ 2 7

Step 2:
The divisor (8) goes into the first digit of the dividend (2), 0 time(s). Therefore, put 0 on top:

0
8 ⟌ 2 7

Step 3:
Multiply the divisor by the result in the previous step (8 x 0 = 0) and write that answer below the dividend.

0
8 ⟌ 2 7
0

Step 4:
Subtract the result in the previous step from the first digit of the dividend (2 - 0 = 2) and write the answer below.

0
8 ⟌ 2 7
- 0
2

Step 5:
Move down the 2nd digit of the dividend (7) like this:

0
8 ⟌ 2 7
- 0
2 7

Step 6:
The divisor (8) goes into the bottom number (27), 3 time(s). Therefore, put 3 on top:

0 3
8 ⟌ 2 7
- 0
2 7

Step 7:
Multiply the divisor by the result in the previous step (8 x 3 = 24) and write that answer at the bottom:

0 3
8 ⟌ 2 7
- 0
2 7
2 4

Step 8:
Subtract the result in the previous step from the number written above it. (27 - 24 = 3) and write the answer at the bottom.

0 3
8 ⟌ 2 7
- 0
2 7
- 2 4
3

You are done, because there are no more digits to move down from the dividend.

The answer is the top number and the remainder is the bottom number.

Therefore, the answer to 27 divided by 8 calculated using Long Division is:

3
8 0
3 years ago
Select three types of lines that engineers use to help represent the shape of a design in a sketch.
Vikki [24]

Hidden lines

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Extension lines:-

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Object lines

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3 years ago
Why some types of aggregate are susceptible to damage from repeated freezing and thawing? Explain.
koban [17]

Answer:

Some types of aggregate are susceptible to damage from repeated freezing and thawing due to their porosity. An aggregate being porous allows water molecules to enter in between the rocks.

When freezing occurs, water is known to expand. The expansion of this in the rocks creates a type of pressure which results in the fracture of the rocks. Subsequent freezing and thawing will allow for more fracture between the rock particles which will lead to its disintegration.

8 0
4 years ago
Which one of these is NOT considered a skill?
Ugo [173]

Answer:

number 2 people skills

Explanation:

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4 0
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Read 2 more answers
Write a program that dynamically allocates an array large enough to hold a user-defined number of test scores. Once all the scor
Readme [11.4K]

Answer:

#include <iostream>

#include <iomanip>

using namespace std;

//Function prototypes

void arrSelectSort(double *, int);

double arrAvgScore(double *, int);

int main()

{  

//Variables  definition

double *TestScores,  

total = 0.0,

average;

int numTest,

count;

//Enter the number of test scores you want to get to their average in ascending order

cout << "How many test scores do you wish to enter?";

cin >> numTest;

//Dynamically allocate an array large enough to hold that many scores

TestScores = new double[numTest];

//Get the test scores

cout << "Enter the test scores below.\n";

for (count = 0; count < numTest; count++)

{

//Display score

cout << "Test Score " << (count + 1) << ": ";

cin >> TestScores[count];

 }  

// Input validation. Only numbers between 0-100

while (numTest<0)

{

cout << "You must enter a scores that non-negative" << endl;

cout << "Please enter a non-negative interger between 0 and 100: ";

cin >> TestScores[count];

}

//Calculate the total test scores

for (count = 0; count < numTest; count++)

{

total += TestScores[count];

}

average = total / numTest;

//Dsiplay the results

cout << fixed << showpoint << setprecision(2);

cout << "The average of all the test score is " << average << endl;

//Free dynamically allocated memory

delete [] TestScores;

TestScores = 0; //make TestScores point to null

//Display the Test Scores in ascending order

cout << "The test scores, sorted in ascending order, are: \n";

system ("pause");

return 0;

}

//Ascending order selection sort

void arrSelectSort(double *arr, int size)

{

int startScan;

double minIndex;

double minElem;  

for(startScan = 0; startScan < (size - 1); startScan++)

{

minIndex = startScan;

minElem = arr[startScan];  

}

for(int index = startScan + 1; index < size; index++)

{

if (arr[index] < minElem)

{

minElem = arr[index];

minIndex = index;  

}

}

void arrAvgScore (double *arr[], int size)

{

double total = 0;

int numTest;

for (int count = 0; count < numTest; count++)

{

total += numTest[count];

average = total / numTest;

}

}

}

6 0
4 years ago
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