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Paha777 [63]
3 years ago
12

The line touching the circle at a point ....................... is known as ........................... .

Engineering
1 answer:
Mashutka [201]3 years ago
8 0
The line touching the circle at a point is known as the radius.
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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
3 years ago
Which rules of the road apply to people riding bicycles, under Illinois law? *
bulgar [2K]

Answer:

C-People biking must follow all rules and laws applicable to a motorist, with some minor exceptions​

Explanation:

People biking should ride on the right side of the right lane when safe, except to pass or make a left turn. When there is only one lane for traffic traveling in each direction and passing is permitted, the center of the street is marked with a broken yellow stripe

~Hope this helps!

4 0
3 years ago
1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per
KiRa [710]

Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

3.

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

4.

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

6.

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

7 0
2 years ago
Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

5 0
3 years ago
How do I do this?<br> Blueprints, complete the missing view.
Ymorist [56]

Explanation:

Look at the drawings and decide which view is missing. Front? Side? Top? Then draw it

7 0
2 years ago
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