Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the
1 answer:
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Answer:
#include <iostream>
using namespace std;
// Pixel structure
struct Pixel
{
unsigned int red;
unsigned int green;
unsigned int blue;
Pixel() {
red = 0;
green = 0;
blue = 0;
}
};
// function prototype
int energy(Pixel** image, int x, int y, int width, int height);
// main function
int main() {
// create array of pixel 3 by 4
Pixel** image = new Pixel*[3];
for (int i = 0; i < 3; i++) {
image[i] = new Pixel[4];
}
// initialize array
image[0][0].red = 255;
image[0][0].green = 101;
image[0][0].blue = 51;
image[1][0].red = 255;
image[1][0].green = 101;
image[1][0].blue = 153;
image[2][0].red = 255;
image[2][0].green = 101;
image[2][0].blue = 255;
image[0][1].red = 255;
image[0][1].green = 153;
image[0][1].blue = 51;
image[1][1].red = 255;
image[1][1].green = 153;
image[1][1].blue = 153;
image[2][1].red = 255;
image[2][1].green = 153;
image[2][1].blue = 255;
image[0][2].red = 255;
image[0][2].green = 203;
image[0][2].blue = 51;
image[1][2].red = 255;
image[1][2].green = 204;
image[1][2].blue = 153;
image[2][2].red = 255;
image[2][2].green = 205;
image[2][2].blue = 255;
image[0][3].red = 255;
image[0][3].green = 255;
image[0][3].blue = 51;
image[1][3].red = 255;
image[1][3].green = 255;
image[1][3].blue = 153;
image[2][3].red = 255;
image[2][3].green = 255;
image[2][3].blue = 255;
// create 3by4 array to store energy of each pixel
int energies[3][4];
// calculate energy for each pixel
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 4; j++) {
energies[i][j] = energy(image, i, j, 3, 4);
}
}
// print energies of each pixel
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 3; j++) {
// print by column
cout << energies[j][i] << " ";
}
cout << endl;
}
}
// function prototype
int energy(Pixel** image, int x, int y, int width, int height) {
// get adjacent pixels
Pixel left, right, up, down;
if (x > 0) {
left = image[x - 1][y];
if (x < width - 1) {
right = image[x + 1][y];
}
else {
right = image[0][y];
}
}
else {
left = image[width - 1][y];
if (x < width - 1) {
right = image[x + 1][y];
}
else {
right = image[0][y];
}
}
if (y > 0) {
up = image[x][y - 1];
if (y < height - 1) {
down = image[x][y + 1];
}
else {
down = image[x][0];
}
}
else {
up = image[x][height - 1];
if (y < height - 1) {
down = image[x][y + 1];
}
else {
down = image[x][0];
}
}
// calculate x-gradient and y-gradient
Pixel x_gradient;
Pixel y_gradient;
x_gradient.blue = right.blue - left.blue;
x_gradient.green = right.green - left.green;
x_gradient.red = right.red - left.red;
y_gradient.blue = down.blue - up.blue;
y_gradient.green = down.green - up.green;
y_gradient.red = down.red - up.red;
int x_value = x_gradient.blue * x_gradient.blue + x_gradient.green * x_gradient.green + x_gradient.red * x_gradient.red;
int y_value = y_gradient.blue * y_gradient.blue + y_gradient.green * y_gradient.green + y_gradient.red * y_gradient.red;
// return energy of pixel
return x_value + y_value;
}
Explanation:
Please see attachment for ouput
To solve this problem we will use the Froude number that relates the Forces of Inertia with the Forces of Gravity. There will be jump in the downstream only if Froude Number (Fr) is greater than 1 at upstream. Our values are given as,
Then the velocity would be:
The number of Froude is given as,
Where,
V = Velocity
g = Gravity
D = Diameter
Replacing we have that
There will be no Jump, correct answer is B.