Answer:
a) D = 4.88 * 10^(-14) m^2 / s
b) <em>t </em>= 1.1 hr
Solution:
a) Magnesium in Aluminum diffusion:
D = Do * exp(-Qd / RT)
= (1.2 * 10^(-4) m^2 / s) * exp ( - 130,000 / 8.31 * 723.15)
= 4.88 * 10^(-14) m^2 / s
D = 4.88 * 10^(-14) m^2 / s
b) The composition remains same at say the distance Xo:
(Cx - Co) / (Cs - Co) = Constant
Xo^2 / D_{550} * <em>t </em>= Xo^2 / D_{450} * 15
D_{550} * <em>t </em>= D_{450} * 15
(1.2 * 10^(-4) m^2 / s) * exp ( - 130,000 / 8.31 * 723.15) * <em>t</em>
(4.88 * 10^(-14) m^2 / s) * 15
by, solving for <em>t </em>we get:
<em>t </em>= 1.1 hr
So, the time required is 1.1 hr.
import java.util.Scanner;
public class JavaApplication58 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter a positive integer:");
String num = scan.nextLine();
for (int i = num.length()-1; i >=0; i--){
System.out.println(num.charAt(i));
}
}
}
I hope this helps!
Answer:
The statement in Python is:
print("The average pH of citrus fruits is ",avg_citrus_pH)
Java
System.out.print("The average pH of citrus fruits is "+avg_citrus_pH);
C++
cout<<"The average pH of citrus fruits is "<<avg_citrus_pH;
Explanation:
The programming language is not stated; so, I answered the question in 3 languages (Python, Java and C++)
Assume that avg_citrus_pH has been declared and initialized; all you need to do is invoke a print statement and then append the variable
In Python, use print()
In c++, use cout<<
In Java, use System.out.print()
So, the statements are:
Python:
print("The average pH of citrus fruits is ",avg_citrus_pH)
Java
System.out.print("The average pH of citrus fruits is "+avg_citrus_pH);
C++
cout<<"The average pH of citrus fruits is "<<avg_citrus_pH;
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In excel spreadsheets, you need to consider specifying relationships between the information you have
stored in your spreadsheets when creating formulas. The elements that will help
you understand using spreadsheet are the constants, operators, references and
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