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2 years ago

How do i do this? if y’all don’t mind helping lol

Engineering

1 answer:

Nookie1986 [14]2 years ago

7 0

Get a ruler and measure it and for every 1/8 of an inch you multiply by 1 to get your footage

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Thoughts about drinking and driving

poizon [28]

Answer: it’s very bad behavior and can be dangerous

Explanation:

8 0

2 years ago

Read 2 more answers

It is the tool used to measure the amount of electric current

sergeinik [125]

Answer:

Ammeter

Explanation:

Instrument for measuring either direct or alternating electric current, in amperes. Ammeters vary in their operating principles and accuracies

7 0

2 years ago

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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

2 years ago

A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2

NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

$\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2$

under ideal condition v₁² = 0

v₂² = 88.88

v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = $\dfrac{\pi}{4}d_1^2 \times v$

Q = $\dfrac{\pi}{4}0.025^2 \times 9.43$

Q= 4.6 × 10⁻³ m³/s

we know that

$C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s$

hence , actual velocity will be equal to 8.39 m/s

6 0

2 years ago

What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin

DochEvi [55]

Answer:

$m_{LP}=0.45\,kg$

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

$Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]$