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uysha [10]
3 years ago
6

What 2 forces move the secondary piston ahead?

Engineering
1 answer:
jekas [21]3 years ago
5 0

Answer:

The primary piston activates one of the two subsystems. The hydraulic pressure created, and the force of the primary piston spring, moves the secondary piston forward.

You might be interested in
Based on the ELR, the stability classification of the atmosphere at the airport between the surface and 3000 m is
Anna11 [10]

Answer:

The answer is "conditionally unstable"

Explanation:

The conditional volatility is really a condition of uncertainty, which reflects on whether increasing air is polluted or not. It determines the rate of ambient delay, which has been between humid and dry adiabatic rates. In general, the environment is in an unilaterally unhealthy region.

Classification dependent on ELR:

Larger than 10 \frac{^{\circ}C}{1000} m Around 10 and 6 \frac{^{\circ}C}{1000} m or less 6 \frac{^{\circ}C}{1000} m volatile implicitly unreliable Therefore ELR is implicitly unreliable 9 \frac{^{\circ}C}{1000} m, that's why it is "conditionally unstable".

6 0
4 years ago
Consider an experiment in which the number of pumps in use at each of two seven-pump gas stations was determined. Call the two g
MakcuM [25]

Answer: Determine the number of pumps in each of the two six-pump gas stations.

• X = the total No. of pumps that are in use at the 2 stations

• Y = the difference between the No. of pumps in utilization at station 1 and the

NO. of pumps in use at the station 2

• U = the max number of pumps in use at the 2 stations

W (observed) = (3, 4)

X (W) = 3 + 4 = 7

Y (W) = 3 − 4 = − 1

U (W) = max (3, 4) = 4

8 0
3 years ago
g a heat engine is located between thermal reservoirs at 400k and 1600k. the heat engine operates with an efficiency that is 70%
Bond [772]

Answer:

<em>Heat rejected to cold body = 3.81 kJ</em>

Explanation:

Temperature of hot thermal reservoir Th = 1600 K

Temperature of cold thermal reservoir Tc = 400 K

<em>efficiency of the Carnot's engine = 1 - </em>\frac{Tc}{Th}<em> </em>

eff. of the Carnot's engine = 1 - \frac{400}{600}

eff = 1 - 0.25 = 0.75

<em>efficiency of the heat engine = 70% of 0.75 = 0.525</em>

work done by heat engine = 2 kJ

<em>eff. of heat engine is gotten as = W/Q</em>

where W = work done by heat engine

Q = heat rejected by heat engine to lower temperature reservoir

from the equation, we can derive that

heat rejected Q = W/eff = 2/0.525 = <em>3.81 kJ</em>

6 0
3 years ago
An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin
Leni [432]

Answer:

Q=41.33 KVAR\ \\at\\\ 480 Vrms

Explanation:

From the question we are told that:

Voltage V=480/0 \textdegree V

Power P=120kW

Initial Power factor p.f_1=0.77 lagging

Final Power factor p.f_2=0.9 lagging

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

p.f_1=0.77

cos \theta_1 =0.77

\theta_1=cos^{-1}0.77

\theta_1=39.65 \textdegree

And

p.f_2=0.9

cos \theta_2 =0.9

\theta_2=cos^{-1}0.9

\theta_2=25.84 \textdegree

Therefore

Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=-41.33VAR

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

Q=41.33 KVAR\ \\at\\\ 480 Vrms

6 0
3 years ago
Implement
kolbaska11 [484]

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

6 0
3 years ago
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