Answer:
See attached file please.
Explanation:
See attached file for detailed explanation and code.
import java.util.*;
class LinklistImplementQueue {
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
/* Creating object of class SLLQueue */
SLLQueue lq = new SLLQueue();
char ch;
do
{
System.out.println("\nQueue Operations");
System.out.println("1. ENQUEUE");
System.out.println("2. DEQUEUE");
int choice = scan.nextInt();
.
.
.
.
See attached file for complete code.
Answer:
the maximum length of specimen before deformation is found to be 235.6 mm
Explanation:
First, we need to find the stress on the cylinder.
Stress = σ = P/A
where,
P = Load = 2000 N
A = Cross-sectional area = πd²/4 = π(0.0037 m)²/4
A = 1.0752 x 10^-5 m²
σ = 2000 N/1.0752 x 10^-5 m²
σ = 186 MPa
Now, we find the strain (∈):
Elastic Modulus = Stress / Strain
E = σ / ∈
∈ = σ / E
∈ = 186 x 10^6 Pa/107 x 10^9 Pa
∈ = 1.74 x 10^-3 mm/mm
Now, we find the original length.
∈ = Elongation/Original Length
Original Length = Elongation/∈
Original Length = 0.41 mm/1.74 x 10^-3
<u>Original Length = 235.6 mm</u>
Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.
Explanation:
Answer:
i believe the law requires you to stop
