Answer:
W = 112 lb
Explanation:
Given:
- δb = 0.025 in
- E = 29000 ksi (A-36)
- Area A_de = 0.002 in^2
Find:
Compute Weight W attached at C
Solution:
- Use proportion to determine δd:
δd/5 = δb/3
δd = (5/3) * 0.025
δd = 0.0417 in
- Compute εde i.e strain in DE:
εde = δd / Lde
εde = 0.0417 / 3*12
εde = 0.00116
- Compute stress in DE, σde:
σde = E*εde
σde = 29000*0.00116
σde = 33.56 ksi
- Compute the Force F_de:
F_de = σde *A_de
F_de = 33.56*0.002
F_de = 0.0672 kips
- Equilibrium conditions apply:
(M)_a = 0
3*W - 5*F_de = 0
W = (5/3)*F_de
W = (5/3)* 0.0672 = 112 lb
Answer:
percentage change in volume = 0.00285 %
Explanation:
given data
bulk modulus = 3.5 × N/m²
bulk stress = N/m²
solution
we will apply here bulk modulus formula that is
bulk modulus = ...............1
put here value and we get
3.5 × =
solve it we get
bulk strain = 2.8571 ×
and
bulk strain =
so that percentage change in volume is = 2.8571 × × 100
percentage change in volume = 0.00285 %
Answer:
See explaination
Explanation:
Code;
import java.util.Scanner;
public class NumberPattern {
public static int x, count;
public static void printNumPattern(int num1, int num2) {
if (num1 > 0 && x == 0) {
System.out.print(num1 + " ");
count++;
printNumPattern(num1 - num2, num2);
} else {
x = 1;
if (count >= 0) {
System.out.print(num1 + " ");
count--;
if (count < 0) {
System.exit(0);
}
printNumPattern(num1 + num2, num2);
}
}
}
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int num1;
int num2;
num1 = scnr.nextInt();
num2 = scnr.nextInt();
printNumPattern(num1, num2);
}
}
See attachment for sample output
