Explain Flags in ARM Processor

If a shear stress acts in one plane of an element, there must be an equal and opposite shear stress acting on a plane that is

xxMikexx [17]

Answer:

90 degrees

Explanation:

In the case when the sheer stress acts in the one plane of an element so it should be equal and opposite also the shear stress acted on a plan i.e. 90 degrees from the plane

Therefore as per the given situation it should be 90 degrees from the plane

hence, the same is to be considered and relevant too

5 0

3 years ago

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Stream Piracy – Kaaterskill, NY. Check and double-click the Problem 15 folder. The dark blue and orange streams highlight the pr

baherus [9]

Answer:

b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.

Explanation:

From the Kaaterskill NY 15 minute map (1906), this shows two classic examples of stream capture.

The Kaaterskill Creek flow down the east relatively steep slopes into the Hudson River Valley. While, the Gooseberry Creek is a low gradient stream flowing down the west direction which in turn drains the higher parts of the Catskills in this area.

However, there is Headward erosion of Kaaterskill Creek which resulted to the capture of part of the headwaters of Gooseberry Creek.

The evidence for this is the presence of "barbed" (enters at obtuse rather than acute angle) tributary which enters Kaaterskill Creek from South Lake which was once a part of the Gooseberry Creek drainage system.

It should be noted again, that there is drainage divide between the Gooseberry and Kaaterskill drainage systems (just to the left of the word Twilight) which is located in the center of the valley.

As it progresses, this divide will then move westward as Kaaterskill captures more and more of the Gooseberry system.

5 0

3 years ago

g For this project you are required to perform Matrix operations (Addition, Subtraction and Multiplication). For each of the ope

Kruka [31]

Answer:

C++ code is explained below

Explanation:

#include<iostream>

using namespace std;

//Function Declarations

void add();

void sub();

void mul();

//Main Code Displays Menu And Take User Input

int main()

{

int choice;

cout << "\nMenu";

cout << "\nChoice 1:addition";

cout << "\nChoice 2:subtraction";

cout << "\nChoice 3:multiplication";

cout << "\nChoice 0:exit";

cout << "\n\nEnter your choice: ";

cin >> choice;

cout << "\n";

switch(choice)

{

case 1: add();

break;

case 2: sub();

break;

case 3: mul();

break;

case 0: cout << "Exited";

exit(1);

default: cout << "Invalid";

}

main();

}

//Addition Of Matrix

void add()

{

int rows1,cols1,i,j,rows2,cols2;

cout << "\nmatrix1 # of rows: ";

cin >> rows1;

cout << "\nmatrix1 # of columns: ";

cin >> cols1;

int m1[rows1][cols1];

//Taking First Matrix

for(i=0;i<rows1;i++)

for(j=0;j<cols1;j++)

{

cout << "\nEnter element (" << i << "," << j << "): ";

cin >> m1[i][j];

cout << "\n";

}

//Printing 1st Matrix

for(i=0;i<rows1;i++)

{

for(j=0;j<cols1;j++)

cout << m1[i][j] << " ";

cout << "\n";

}

cout << "\nmatrix2 # of rows: ";

cin >> rows2;

cout << "\nmatrix2 # of columns: ";

cin >> cols2;

int m2[rows2][cols2];

//Taking Second Matrix

for(i=0;i<rows2;i++)

for(j=0;j<cols2;j++)

{

cout << "\nEnter element (" << i << "," << j << "): ";

cin >> m2[i][j];

cout << "\n";

}

//Displaying second Matrix

cout << "\n";

for(i=0;i<rows2;i++)

{

for(j=0;j<cols2;j++)

cout << m2[i][j] << " ";

cout << "\n";

}

//Displaying Sum of m1 & m2

if(rows1 == rows2 && cols1 == cols2)

{

cout << "\n";

for(i=0;i<rows1;i++)

{

for(j=0;j<cols1;j++)

cout << m1[i][j]+m2[i][j] << " ";

cout << "\n";

}

else

cout << "operation is not supported";

main();

}

void sub()

{

int rows1,cols1,i,j,k,rows2,cols2;

cout << "\nmatrix1 # of rows: ";

cin >> rows1;

cout << "\nmatrix1 # of columns: ";

cin >> cols1;

int m1[rows1][cols1];

for(i=0;i<rows1;i++)

for(j=0;j<cols1;j++)

{

cout << "\nEnter element (" << i << "," << j << "): ";

cin >> m1[i][j];

cout << "\n";

}

for(i=0;i<rows1;i++)

{

for(j=0;j<cols1;j++)

cout << m1[i][j] << " ";

cout << "\n";

}

cout << "\nmatrix2 # of rows: ";

cin >> rows2;

cout << "\nmatrix2 # of columns: ";

cin >> cols2;

int m2[rows2][cols2];

for(i=0;i<rows2;i++)

for(j=0;j<cols2;j++)

{

cout << "\nEnter element (" << i << "," << j << "): ";

cin >> m2[i][j];

cout << "\n";

}

for(i=0;i<rows2;i++)

{

for(j=0;j<cols2;j++)

cout << m1[i][j] << " ";

cout << "\n";

}

cout << "\n";

//Displaying Subtraction of m1 & m2

if(rows1 == rows2 && cols1 == cols2)

{

for(i=0;i<rows1;i++)

{

for(j=0;j<cols1;j++)

cout << m1[i][j]-m2[i][j] << " ";

cout << "\n";

}

else

cout << "operation is not supported";

main();

}

void mul()

{

int rows1,cols1,i,j,k,rows2,cols2,mul[10][10];

cout << "\nmatrix1 # of rows: ";

cin >> rows1;

cout << "\nmatrix1 # of columns: ";

cin >> cols1;

int m1[rows1][cols1];

for(i=0;i<rows1;i++)

for(j=0;j<cols1;j++)

{

cout << "\nEnter element (" << i << "," << j << "): ";

cin >> m1[i][j];

cout << "\n";

}

cout << "\n";

for(i=0;i<rows1;i++)

{

for(j=0;j<cols1;j++)

cout << m1[i][j] << " ";

cout << "\n";

}

cout << "\nmatrix2 # of rows: ";

cin >> rows2;

cout << "\nmatrix2 # of columns: ";

cin >> cols2;

int m2[rows2][cols2];

for(i=0;i<rows2;i++)

for(j=0;j<cols2;j++)

{

cout << "\nEnter element (" << i << "," << j << "): ";

cin >> m2[i][j];

cout << "\n";

}

cout << "\n";

//Displaying Matrix 2

for(i=0;i<rows2;i++)

{

for(j=0;j<cols2;j++)

cout << m2[i][j] << " ";

cout << "\n";

}

if(cols1!=rows2)

cout << "operation is not supported";

else

{

//Initializing results as 0

for(i = 0; i < rows1; ++i)

for(j = 0; j < cols2; ++j)

mul[i][j]=0;

// Multiplying matrix m1 and m2 and storing in array mul.

for(i = 0; i < rows1; i++)

for(j = 0; j < cols2; j++)

for(k = 0; k < cols1; k++)

mul[i][j] += m1[i][k] * m2[k][j];

// Displaying the result.

cout << "\n";

for(i = 0; i < rows1; ++i)

for(j = 0; j < cols2; ++j)

{

cout << " " << mul[i][j];

if(j == cols2-1)

cout << endl;

}

main();

}

5 0

3 years ago

Use the graph to determine which statement is true about the end behavior of f(x).

Airida [17]

Answer:

As the x-values go to negative infinity, the function’s values go to positive infinity.

Explanation:

if the ans choices are:

As the x-values go to negative infinity, the function’s values go to negative infinity.

As the x-values go to negative infinity, the function’s values go to positive infinity.

As the x-values go to positive infinity, the function’s values go to negative infinity.

As the x-values go to positive infinity, the function’s values go to zero.

the ans is the 2nd choice

4 0

3 years ago

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An L2 steel strap having a thickness of 0.125 in. and a width of 2 in. is bent into a circular arc of radius 600 in. Determine t

lesya692 [45]

Answer:

the maximum bending stress in the strap is 3.02 ksi

Explanation:

Given the data in the question;

steel strap thickness = 0.125 in

width = 2 in

circular arc radius = 600 in

we know that, standard value of modulus of elasticity of L2 steel is; E = 29 × 10³ ksi;

Now, using simple theory of bending

1/p = M/EI

solve for M

Mp = EI

M = EI / p ----- let this be equation 1

The maximum bending stress in the strap is;

σ = Mc / I -------let this be equation 2

substitute equation 1 into 2

σ = ( EI / p)c / I

σ = ( c/p )E

so we substitute in our values

σ = ( (0.125/2) / 600 )29 × 10³

σ = 0.00010416666 × 29 × 10³

σ = 3.02 ksi

Therefore, the maximum bending stress in the strap is 3.02 ksi

3 0

3 years ago

Explain Flags in ARM Processor​

Explain Flags in ARM Processor