Question
Determine the average water exit velocity
Answer:
53.05 m/s
Explanation:
Given information
Volume flow rate, ![Q=16 m^{3}/min](https://tex.z-dn.net/?f=Q%3D16%20m%5E%7B3%7D%2Fmin)
Diameter d= 8cm= 0.08 m
Assumptions
- The flow is jet flow hence momentum-flux correction factor is unity
- Gravitational force is not considered
- The flow is steady, frictionless and incompressible
- Water is discharged to the atmosphere hence pressure is ignored
We know that Q=AV and making v the subject then
where V is the exit velocity and A is area
Area,
where d is the diameter
By substitution
![V=\frac {16\times 4}{\pi 0.08^{2}}=3183.098862 m/min](https://tex.z-dn.net/?f=V%3D%5Cfrac%20%7B16%5Ctimes%204%7D%7B%5Cpi%200.08%5E%7B2%7D%7D%3D3183.098862%20m%2Fmin)
To convert v to m/s from m/s, we simply divide it by 60 hence
![V=\frac {3183.098862 m/min}{60 s}=53.0516477 m/s\approx 53.05 m/s](https://tex.z-dn.net/?f=V%3D%5Cfrac%20%7B3183.098862%20%20m%2Fmin%7D%7B60%20s%7D%3D53.0516477%20m%2Fs%5Capprox%2053.05%20m%2Fs)
Answer:
class TriangleNumbers
{
public static void main (String[] args)
{
for (int number = 1; number <= 10; ++number) {
int sum = 1;
System.out.print("1");
for (int summed = 2; summed <= number; ++summed) {
sum += summed;
System.out.print(" + " + Integer.toString(summed));
}
System.out.print(" = " + Integer.toString(sum) + '\n');
}
}
}
Explanation:
We need to run the code for each of the 10 lines. Each time we sum numbers from 1 to n. We start with 1, then add numbers from 2 to n (and print the operation). At the end, we always print the equals sign, the sum and a newline character.
Answer:
To be able to develop as a human and give back to the community. To ensure that people have safe food and water to drink, clean fuel and energy sources, and in general a safe enviornment to live in.
Explanation:
Answer:
b) The null hypothesis should be rejected.
Explanation:
The null hypothesis is that the mean shear strength of spot welds is at least
3.1 MPa
H0: u ≥3.1 MPa against the claim Ha: u< 3.1 MPa
The alternate hypothesis is that the mean shear strength of spot welds is less than 3.1 MPa.
This is one tailed test
The critical region Z(0.05) < ± 1.645
The Sample mean= x`= 3.07
The number of welds= n= 15
Standard Deviation= s= 0.069
Applying z test
z= x`-u/s/√n
z= 3.07-3.1/0.069/√15
z= -0.03/0.0178
z= -1.68
As the calculated z= -1.68 falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa