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Dmitry_Shevchenko [17]
2 years ago
10

Write down the understanding of these designations of the materials (a)Ck45, (b)AISI 4140

Engineering
1 answer:
marin [14]2 years ago
7 0

Answer:Don't know sorry

Explanation:

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What is an air mass?​
kotegsom [21]

Answer:

An air mass is a body of air with horizontally uniform temperature, humidity, and pressure.

Explanation:

Because it is

8 0
3 years ago
Read 2 more answers
A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
kompoz [17]

Answer:

a. The mass flow rate (in lbm/s) is 135lbm/s

b. The temperature (in o F) is 200.8°F

Explanation:

We assume that potential energy and kinetic energy are negligible and the control volume operates at a steady state.

Given

a. The mass flow rate (in lbm/s) is 135lbm/s

b.

m1 = Rate at inlet 1 = 125lbm/s

m2 = Rate at inlet 2 = 10lbm/s

The mass flow rate (in lbm/s) is calculated as m1 + m2

Mass flow rate = 125lbm/s + 10lbm/s

Mass flow rate = 135lbm/s

Hence, the mass flow rate (in lbm/s) is 135lbm/s

b. To calculate the temperature.

First we need to determine the enthalpy h1 at 14.7psia

Using table A-3E (thermodynamics)

h1 = 180.15 Btu/Ibm

h2 at 14.7psia and 60°F = 28.08 Btu/Ibm

Calculating h3 using the following formula

h3 = (h1m1 + h2m2) / M3

h3 = (180.15 * 125 + 28.08 * 10)/135

h3 = 168.8855555555555

h3 = 168.89 Btu/Ibm

To get the final temperature; we make use of table A-2E of thermodynamics.

Because h3 < h1, it means the liquid is at a compressed state.

The corresponding temperature at h3 = 168.89 is 200.8°F

The temperature (in o F) is 200.8°F

6 0
3 years ago
13. Write a function which is passed two strings. The function creates a new string from the two original strings by copying one
attashe74 [19]

Answer:

I am writing the code in C++. Let me know if you want the program in some other programming language.

#include <iostream>  // includes header file for input output functions

using namespace std;     //to identify objects like cin cout

string CopyStrings(string string1, string string2)  

{   string newString = "";    

   for (int loop = 0; loop < string1.length() ||  

                   loop < string2.length(); loop++)      {      

       if (loop < string1.length())  

           newString += string1[loop];          

       if (loop < string2.length())  

           newString += string2[loop];      }  

   return newString;   }  

int main()  

{   string stringA = "ace";  

   string stringB = "bdf";  

   cout << CopyStrings(stringA, stringB);   }

Output:

abcdef

Explanation:

The function CopyStrings() function takes two strings i.e. string1 and string2 as parameters to copy characters from both the string one character from each.

The newString variable stores the new string after copying characters from both strings string1 and string2.

Then the for loop starts which has a variable loop which is an index variable that traverses through both the strings stored in string1 and string2. The loop continues to execute until it moves through entire length of string1 and string2 which means it copies all the characters from both string1 and string2. length() is used here which returns length of the string1 and string2.

If statement in the for loop checks the character that loop (index) variable is pointing to is less than the string1 length which means it checks each character stored in string1. For example if string1 contains "ace" and loop variable is moving through the string and is currently at "a" then this condition is true. If the condition evaluates to true then the body of if statement is executed. The next statement stores that character a into the newString variable.

Next If statement checks character that loop variable is pointing to is less than the string2 length which means it checks each character stored in string2. For example if string2 contains "bdf" and loop variable is moving through the string and is currently at "b" then this condition is true. If the condition evaluates to true then the body of if statement is executed. The next statement stores that character b into the newString variable.

Then the second iteration starts which again first stores the next character i.e. c from string1 into newString and then stores next character i.e d from string2 into newString.

Then the third iteration starts which again first stores the next character i.e. e from string1 into newString and then stores next character i.e f from string2 into newString.

Then the loop breaks as the loop variable reaches end of both the string1 and string2.

return newString will return the copied string into the output screen which is abcdef.

The screenshot of code along with output is attached.

3 0
3 years ago
Unit for trigonometric functions is always "radian". 1. 10 points: Do NOT submit your MATLAB code for this problem (a) Given f(x
RoseWind [281]

Answer:

Below is the required code.

Explanation:

%% Newton Raphson Method

clear all;

clc;

x0=input('Initial guess:\n');

x=x0;

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=10;

i=0;

cc=input('Condition of convergence:\n');

while ep>=cc

i=i+1;

temp=x;

x=x-(f/g);

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=abs(x-temp);

fprintf('x = %6f and error = %6f at iteration = %2f \n',x,ep,i);

end

fprintf('The solution x = %6f \n',x);

%% End of MATLAB Program

Command Window:

(a) First Root:

Initial guess:

1.5

Condition of convergence:

0.01

x = -1.815662 and error = 3.315662 at iteration = 1.000000

x = -0.644115 and error = 1.171547 at iteration = 2.000000

x = 0.208270 and error = 0.852385 at iteration = 3.000000

x = 0.434602 and error = 0.226332 at iteration = 4.000000

x = 0.451631 and error = 0.017029 at iteration = 5.000000

x = 0.451732 and error = 0.000101 at iteration = 6.000000

The solution x = 0.451732

>>

Second Root:

Initial guess:

3.5

Condition of convergence:

0.01

x = 3.300299 and error = 0.199701 at iteration = 1.000000

x = 3.305650 and error = 0.005351 at iteration = 2.000000

The solution x = 3.305650

>>

(b) Guess x=0.5:

Initial guess:

0.5

Condition of convergence:

0.01

x = 0.450883 and error = 0.049117 at iteration = 1.000000

x = 0.451732 and error = 0.000849 at iteration = 2.000000

The solution x = 0.451732

>>

Guess x=1.75:

Initial guess:

1.75

Condition of convergence:

0.01

x = 227.641471 and error = 225.891471 at iteration = 1.000000

x = 218.000998 and error = 9.640473 at iteration = 2.000000

x = 215.771507 and error = 2.229491 at iteration = 3.000000

x = 217.692636 and error = 1.921130 at iteration = 4.000000

x = 216.703197 and error = 0.989439 at iteration = 5.000000

x = 216.970438 and error = 0.267241 at iteration = 6.000000

x = 216.971251 and error = 0.000813 at iteration = 7.000000

The solution x = 216.971251

>>

Guess x=3.0:

Initial guess:

3

Condition of convergence:

0.01

x = 3.309861 and error = 0.309861 at iteration = 1.000000

x = 3.305651 and error = 0.004210 at iteration = 2.000000

The solution x = 3.305651

>>

Guess x=4.7:

Initial guess:

4.7

Condition of convergence:

0.01

x = -1.916100 and error = 1.051861 at iteration = 240.000000

x = -0.748896 and error = 1.167204 at iteration = 241.000000

x = 0.162730 and error = 0.911626 at iteration = 242.000000

x = 0.428332 and error = 0.265602 at iteration = 243.000000

x = 0.451545 and error = 0.023212 at iteration = 244.000000

x = 0.451732 and error = 0.000187 at iteration = 245.000000

The solution x = 0.451732

>>

Explanation:

The two solutions are x =0.451732 and 3.305651 within the range 0 < x< 5.

The initial guess x = 1.75 fails to determine the solution as it's not in the range. So the solution turns to unstable with initial guess x = 1.75.

7 0
3 years ago
we wish to send at a rate of 10Mbits/s over a passband channel. Assuming that an excess bandwidth of 50% is used, how much bandw
gayaneshka [121]

Answer:

QPSK: 7.5 MHz

64-QAM:2.5 MHz

64-Walsh-Hadamard: 160 MHz

Explanation:

See attached picture.

6 0
3 years ago
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