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umka21 [38]
3 years ago
11

Write a function call with arguments tensPlace, onesPlace, and userInt. Be sure to pass the first two arguments as pointers. Sam

ple output for the given program: tensPlace = 4, onesPlace = 1
Engineering
1 answer:
Trava [24]3 years ago
4 0

Answer:

#include <stdio.h>

void SplitIntoTensOnes(int* tensDigit, int* onesDigit, int DecVal){

  *tensDigit = (DecVal / 10) % 10;

  *onesDigit = DecVal % 10;

  return;

}

int main(void) {

  int tensPlace = 0;

  int onesPlace = 0;

  int userInt = 0;

  userInt = 41;

  SplitIntoTensOnes(&tensPlace, &onesPlace, userInt);

  printf("tensPlace = %d, onesPlace = %d\n", tensPlace, onesPlace);

  return 0;

}

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Explain the Corona discharge?​
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Answer:

• Corona discharge means <u> </u><u>neutralization</u><u> </u><u>of</u><u> </u><u>charge</u><u> </u><u>action</u><u> </u><u>at</u><u> </u><u>a</u><u> </u><u>sharp</u><u> </u><u>point</u>

example; a pear shaped conductor, lightening conductor, van de graaf generator.

Explanation:

• In corona discharge, When a sharp conductor is placed somewhere in open space. The charges of opposite chrage attract the opposite ions in the air around that conductor.

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3 0
3 years ago
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jenyasd209 [6]

Answer:

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3 years ago
Read 2 more answers
Mercury flows inside a copper tube 9m long with a 5.1сm inside diameter at an average velocity of 7.0 m/s. The inside surface te
svp [43]

Answer:

rate of heat transfer = 9085708.80 W

Explanation:

Given:

Inside diameter, D = 5.1 cm

                               = 5.1 x 10^{-2} m

Average velocity, V = 7 m/s

Mean temperature, T = (66+38) /2

                                    = 52°C

Therefore kinematic viscosity at 52°C is ν = 0.104 X 10^{-6} m^{2} / s

Prandtl no., Pr = 0.021

We know Renold No. is

Re = \frac{V\times D}{\nu }

Re = \frac{7\times 5.1\times 10^{-2}}{0.104\times 10^{-6}}

     = 3.432 X 10^{6}

Therefore the flow is turbulent.

Since the flow is turbulent and the ratio of L/D is greater than 60 we can use Dittua-Boelter equation.

Nu = 0.023 Re^{0.8}.Pr^{0.3}

     = 0.023 x (3.432 \times10^{6})^{0.8} x (0.021)^{0.3}

     = 1221.52

Since Nu = \frac{h.D}{k}

          h = \frac{k\times Nu}{D}

             = \frac{9.4\times 1221.52}{5.1\times 10^{-2}}

             = 225143.3

Therefore rate of heat transfer, q = h.A(T-T_{\infty }

           q= 225143.3 x 2πrh ( 66-38)

             = 225143.3 X 2π X \frac{5.1\times10^{-2}}{2}\times 9\times 28

              = 9085708.80 W

6 0
3 years ago
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