100: D, third law of motion
101: D, second law of motion
Answer:
Heat losses by convection, Qconv = 90W
Heat losses by radiation, Qrad = 5.814W
Explanation:
Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:
1. Radiation
2. Conduction
3. Convection
Convection is defined as the transfer of heat through the actual movement of the molecules.
Qconv = hA(Temp.final - Temp.surr)
Where h = 6.4KW/m2K
A, area of a square = L2
= (0.25)2
= 0.0625m2
Temp.final = 250°C
Temp.surr = 25°C
Q = 64 * 0.0625 * (250 - 25)
= 90W
Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.
Qrad = E*S*(Temp.final4 - Temp.surr4)
Where E = emissivity of the surface
S = boltzmann constant
= 5.6703 x 10-8 W/m2K4
Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)
= 5.814 W
Answer:
Explanation:
Steam at outlet is an superheated steam, since 
. From steam tables, the specific enthalpy is:
The throttle valve is modelled after the First Law of Thermodynamics:
Hence, specific enthalpy at inlet is:
The quality in the steam line is:
Answer:
The level of the service is loss and the density is 34.2248 pc/mi/ln
Explanation:
the solution is attached in the Word file
Answer:
import java.util.*;
public class BarChart
{
public static void main(String args[])
{
int arr[]=new int[5];
Scanner sc=new Scanner(System.in);
for(int i=0;i<5;i++)
{
while(true){
System.out.println("Enter today's sale for store "+(i+1)+" (negative value not allowed)");
arr[i]=sc.nextInt();
if(arr[i]>0)
break;
}
}
System.out.println("SALES BAR CHART");
for(int i=0;i<5;i++)
{
System.out.println("Store "+(i+1)+": ");
for(int j=0;j<arr[i];j=j+100)
{
System.out.print("*");
}
System.out.println("");
}
}
}
