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2 years ago

Which examples best demonstrate likely tasks for Health, Safety, and Environmental Assurance workers? check all that apply

Engineering

2 answers:

Over [174]2 years ago

6 0

Sam, Elijah, Joy Those are 100% correct from my human knowledge

Ira Lisetskai [31]2 years ago

6 0

Answer:

i just did the thing on edge

Explanation:

You might be interested in

Why do engineers play a variety of roles in the engineering process?

Crazy boy [7]

Answer: The engineering design process emphasizes open-ended problem solving and encourages students to learn from failure. This process nurtures students abilities to create innovative solutions to challenges in any subject! In addition to their involvement in design and development, many engineers work in testing, production, or maintenance. These engineers supervise production in factories, determine the causes of a component's failure, and test manufactured products to maintain quality.

Explanation:

6 0

2 years ago

Open the"stateData3.c" program and try to understand how the tokenization works. If you open the input file "stateData.txt", you

babymother [125]

Answer:

Kindly see explaination

Explanation:

Code

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#define size 200

int main(void)

{

int const numStates = 50;

char tempBuffer[size];

char tmp[size];

char fileName[] = "stateData.txt"; // Name of the text file (input file) which contains states and its populations

char outFile[] = "stateDataOutput1.txt"; // Output file name

// Open the input file, quit if it fails...

FILE *instream = fopen(fileName, "r");

/* Output File variable */

FILE *opstream;

if(instream == NULL) {

fprintf(stderr, "Unable to open file: %s\n", fileName);

exit(1);

}

//TODO: Open the output file in write ("w") mode

/* Opening output file in write mode */

opstream = fopen(outFile, "w");

//TODO: Read the file, line by line and write each line into the output file

//Reading data from file

while(fgets(tmp, size, instream) != NULL)

{

//Writing data to file

fputs(tmp, opstream);

}

// Close the input file

fclose(instream);

//TODO: Close the output file

/* Closing output file */

fclose(opstream);

return 0;

}

5 0

3 years ago

Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value

masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

$q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} } )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}$

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

$q_{n} =uK(\frac{g(\rho_{L}-\rho _{v}) }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4$

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0

3 years ago

A distribution center is used in which of the following applications?

FrozenT [24]

both b and c are the right

5 0

3 years ago

Express (118)10 and (-49)10 in 8-bit binary one’s complement form and then add the numbers. What would be the representation (-0

stealth61 [152]

Answer:

$118_{10}= 0110111$ 2) $-49_{10}=110001_{2}$ 3) $0_{10}=0:16 \Rightarrow 0_{10}=0_{16}$

Explanation:

1) Expressing the Division as the summation of the quotient and the remainder

for

118, knowing it is originally a decimal form:

118:2=59 +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1

$118_{10}= 0110111$

2) $-49_{10}$

Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.

|-49|=49

49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1

100011

$-49_{10}=110001_{2}$

3) $(-0)_{10}$

The first step on that is dividing by 16, and then dividing their quotient again by 16, so on and adding their remainders. Simply put:

$(-0)_{10}=0:16=0 \Rightarrow (0)_{10}=0_{16} \:or\\(0)_{16}=0000000000000000$

5 0

3 years ago