Answer: Attached below is the well written question and solution
answer: 
i) Attached below 
ii) similar parameter =  
Explanation:
Using ;  L as characteristic length and Vo as reference velocity 
i) Nondimensionalize the equations
ii) Identifying similarity parameters 
the similar parameters are  = 
Attached below is the detailed solution 
 
        
             
        
        
        
Answer:
import java.util.Scanner;
public class FindMatchValue {
   public static void main (String [] args) {
      Scanner scnr = new Scanner(System.in);
      final int NUM_VALS = 4;
      int[] userValues = new int[NUM_VALS];
      int i;
      int matchValue;
      int numMatches = -99; // Assign numMatches with 0 before your for loop
      matchValue = scnr.nextInt();
      for (i = 0; i < userValues.length; ++i) {
         userValues[i] = scnr.nextInt();
      }
      /* Your solution goes here */
          numMatches = 0;
      for (i = 0; i < userValues.length; ++i) {
         if(userValues[i] == matchValue) {
                        numMatches++;
                 }
      }
      System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);
   }
}
 
        
             
        
        
        
Answer:
power developed by the turbine = 6927.415 kW
Explanation:
given data 
pressure = 4 MPa
specific enthalpy h1 = 3015.4 kJ/kg
velocity v1 = 10 m/s
pressure = 0.07 MPa
specific enthalpy h2 = 2431.7 kJ/kg
velocity v2 = 90 m/s
mass flow rate = 11.95 kg/s
solution
we apply here  thermodynamic equation that 
energy equation that is 

put here value with 
turbine is insulated so q = 0
so here 

solve we get 
w = 579700 J/kg = 579.7 kJ/kg
and
W = mass flow rate × w
W = 11.95 × 579.7
W = 6927.415 kW
power developed by the turbine = 6927.415 kW
 
        
             
        
        
        
Answer:
The amplitude of the absorbed mass can be found
for ka:

now
![w^2=\frac{K_{a} }{m_{a} } \\m_{a} =\frac{K_{a} }{w^2} =\frac{125000}{[6000*2\pi /60]^2} =0.317kg](https://tex.z-dn.net/?f=w%5E2%3D%5Cfrac%7BK_%7Ba%7D%20%7D%7Bm_%7Ba%7D%20%7D%20%5C%5Cm_%7Ba%7D%20%3D%5Cfrac%7BK_%7Ba%7D%20%7D%7Bw%5E2%7D%20%3D%5Cfrac%7B125000%7D%7B%5B6000%2A2%5Cpi%20%2F60%5D%5E2%7D%20%3D0.317kg)
 
        
             
        
        
        
Any point on earth can be located by specifying its latitude and longitude, including Washington, DC, which is pictured here. Lines of latitude and longitude form an imaginary global grid system, shown in Fig. 1.17. Any point on the globe can be located exactly by specifying its latitude and longitude.