Answer:
import java.util.Scanner;
public class InputExample {
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int birthMonth;
int birthYear;
birthMonth = scnr.nextInt();
birthYear = scnr.nextInt();
System.out.println(birthMonth+"/"+birthYear);
}
}
Answer:
R = 31.9 x 10^(6) At/Wb
So option A is correct
Explanation:
Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A
Thus; R = L/μA,
Now from the question,
L = 4m
r_1 = 1.75cm = 0.0175m
r_2 = 2.2cm = 0.022m
So Area will be A_2 - A_1
Thus = π(r_2)² - π(r_1)²
A = π(0.0225)² - π(0.0175)²
A = π[0.0002]
A = 6.28 x 10^(-4) m²
We are given that;
L = 4m
μ_steel = 2 x 10^(-4) Wb/At - m
Thus, reluctance is calculated as;
R = 4/(2 x 10^(-4) x 6.28x 10^(-4))
R = 0.319 x 10^(8) At/Wb
R = 31.9 x 10^(6) At/Wb
Answer:
If the heat engine operates for one hour:
a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.
b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.
In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.
Explanation:
The Carnot efficiency is obtained as:

Where
is the atmospheric temperature and
is the maximum burn temperature.
For the case (B), the efficiency we will use is:

The work done by the engine can be calculated as:
where Hv is the heat value.
If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.

If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.

Answer:
a) 4 passes are required to sort the string.
b) 4
c) i) TARP
ii) CHIP
iii) PART
iv) TARP
v) TARP
d) O(k+n), n is no. of strings, k is largest no. of character in among the string
O(d*(n+10)), n is no. of integers
Explanation:
Answer:
Option A - fail/ not fail
Explanation:
For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _fail______ and using the Maximum Shear Stress Theory the material will ___not fail_______