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Arisa [49]
3 years ago
8

8.2.1: Function pass by reference: Transforming coordinates. Define a function CoordTransform() that transforms the function's f

irst two input parameters xVal and yVal into two output parameters xValNew and yValNew. The function returns void. The transformation is new
Engineering
1 answer:
ahrayia [7]3 years ago
4 0

Answer:

<em>The output will be (3, 4) becomes (8, 10) </em>

Explanation:

#include <stdio.h>

<em>//If you send a pointer to a int, you are allowing the contents of that int to change. </em>

void CoordTransform(int xVal,int yVal,int* xNew,int* yNew){

*xNew = (xVal+1)*2;

*yNew = (yVal+1)*2;

}

int main(void) {

int xValNew = 0;

int yValNew = 0;

CoordTransform(3, 4, &xValNew, &yValNew);

printf("(3, 4) becomes (%d, %d)\n", xValNew, yValNew);

return 0;

}

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The pressure in a water line is 1500 kPa. What is the line pressure in (a) lb/ft2units and (b) lbf/in2(psi) units?
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Answer:

Part A:

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Part B:

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Explanation:

Part A:

Line Pressure is 1500 KPa

We need a conversion factor which converts KPa to lb/ft^2.

20.88543 \frac{lb}{ft^2}= 1\ KPa

In order to convert 1500 KPa to lb/ft^2, we proceed as:

1\ KPa=20.88543 \frac{lb}{ft^2} \\1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa}\\1500\ KPa= 31328.145 \frac{lb}{ft^2}

1500 KPa is 31328.145 lb/ft^2

Part B:

We will use the same procedure we did in Part A:

1 ft= 12 in

(1\ ft)^2=(12\ in)^2\\1 ft^2=144 in^2

Converting 1500 KPa\  into\  \frac{lb}{in^2}

1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa} * \frac{ft^2}{144\ in^2}

1500 KPa=217.55656 \frac{lb}{in^2}(Psi)

1500 KPa is 217.55656 lb/in^2 (psi)

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