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kifflom [539]
3 years ago
8

Yo where are the answer :/

Engineering
1 answer:
Sergeeva-Olga [200]3 years ago
4 0
The answer is in your mind
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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

Diameter D= 1 in

Speed V= 105 sfpm

feed f= 0.015 ipr

Now  the metal   removal  rate   given as

MRR= 12 f V d

d= depth of cut

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

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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
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It is a gun right am I right
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A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.
denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is 8.621 \times 10^{-4} in/in

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

\frac{M}{I}=\frac{F}{y}

Here

  • M is the moment which is to be calculated
  • I is the moment of inertia given as

                         I=\frac{bd^3}{12}

Here

  • b is the breath given as 0.75"
  • d is the depth which is given as 8"

                     I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4

  • y is given as

                     y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\

  • Force is 50 ksi

\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in

The yield moment is 400 k.in.

Part b:

The strain is given as

Strain=\frac{Stress}{Elastic Modulus}

The stress at the station 2" down from the top is estimated by ratio of triangles as

                        F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi

Now the steel has the elastic modulus of E=29000 ksi

Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in

So the strain is 8.621 \times 10^{-4} in/in

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi

The plastic moment is 600 ksi.

3 0
4 years ago
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