Answer:
the surface heat-transfer coefficient due to natural convection during the initial cooling period. = 4.93 w/m²k
Explanation:
check attachement for answer explanation
Consider a cubic workpiece of rigid perfect plastic material with side length lo. The cube is deformed plastically to the shape
of a rectangular parallelepiped of dimensions l1,l2, and l3. Show that the volume constancy requires that the following expression be satisfied: e1+e2+e3 = 0
1 answer:
Answer: ε₁+ε₂+ε₃ = 0
Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-
l₀l₀l₀=l₁l₂l₃
taking natural log on both sides
Considering the logarithmic Laws of division and multiplication :
ln(AB) = ln(A)+ln(B)
ln(A/B) = ln(A)-ln(B)
Use the image attached to see the definition of true strain defined as
ln(l1/1o)= ε₁
which then proves that ε₁+ε₂+ε₃ = 0
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Answer:
change in storage = -310,500 ft^3
intital storage= 3.67 acre ft
Explanation:
Given data:
Rate of inflow = 350 cfs
Rate of outflow = 285 cfs
After 90 min, rate of inflow = 250 cfs
Rate of outflow = 200 cfs
final storage = 10.8 acre-ft
calculating the average inflow and outflow
average inflow
average outlow
total amount of water drain during the period of one hour
= (average outflow - average inflow) *60*90
= (242.5 - 300)*60*90 = -310,500
change in storage is calculate as
= -310,500 ft^3
in cubic meter
= -310500/35.315 = 8792.30 cm^3
in acre-ft
= -310,500/43560 = 7.13 acre ft
initial storage = 10.8 - 7.13 = 3.67 acre ft
Answer:
import java.util.Scanner;
import java.io.*;
public class Main{
public static void main(String []args){
StringBuilder reversedName = new StringBuilder();
StringBuilder subName = new StringBuilder();
Scanner myObj = new Scanner(System.in); // Create a Scanner object
System.out.println("Enter your full name (last name + first name)");
String name = myObj.nextLine(); // Read user input
for (int i = 0; i < name.length(); i++) {
char currentChar = name.charAt(i);
if (currentChar != ' ') {
subName.append(currentChar);
} else {
reversedName.insert(0, currentChar + subName.toString());
subName.setLength(0);
}
}
String Resul = reversedName.insert(0, subName.toString()).toString();
System.out.println("Reverse full name: " + Resul);
}
}
Explanation:
To solve the problem we used StringBuilder, that allow us to build a string letter by letter, we are using two, one to store temporary information (subName) and the other to store the surname plus name (reversedName). After defining the variables, the program asks for the user fullname using Scanner class and it's method .nextLine (the variable 'name' will save the user input). A while loop is also used to iterate over the name, taking character by character with the method .charAt, the program will append letters in subName until it finds a space only then the content of subName will be transfer to reversedName and the content of subName will be deleted, then the program continues with the process until the end of the for loop; finally, the program places the content of subName at the beginning of reversedName, and print the result transforming the StringBuilder to a normal string.
