Answer:
see explaination
Explanation:
import java.util.InputMismatchException;
import java.util.Scanner;
public class calculate {
static float a=0,b=0;
double cal()
{
if(a==0||b==0)
{
System.out.println("no values found in a or b");
start();
}
double x=(a*a)+(b*b);
double h=Math.sqrt(x);
a=0;
b=0;
return h;
}
float enter()
{
float val=0;
try
{
System.out.println("Enter side");
Scanner sc1 = new Scanner(System.in);
val = sc1.nextFloat();
return val;
}
catch(InputMismatchException e)
{
System.out.println("Enter correct value");
}
return val;
}
void start()
{
calculate c=new calculate();
while(true)
{
System.out.println("Enter Command");
Scanner sc = new Scanner(System.in);
String input = sc.nextLine();
switch(input)
{
case "A":
a=c.enter();
break;
case "B":
b=c.enter();
break;
case "C":
double res=c.cal();
System.out.println("Hypotenuse is : "+res);
break;
case "Q":
System.exit(0);
default:System.out.println("wrong command");
}
}
}
public static void main(String[] args) {
calculate c=new calculate();
c.start();
}
}
Answer:
Explanation:
Force on Q₃ due to charge Q₁
= 9 x 10⁹x 4 x 10⁻⁹ x1 x 10⁻⁹ / 5²
= 1.44 x 10⁻⁹ N
Force due to Q₂ will also be 1.44 x 10⁻⁹ N
component of these forces along x axis
-= 2 x 1.44 x 10⁻⁹ cosθ
= 2.88 x 10⁻⁹ x 4/5
= 2.30x10⁻⁹ N along x axis.
The y-component will calcel out.
b ) In this case , Q₁ will repel and Q₂ will attract.
In this case Q₁ will repel and Q₂ will attract. Component along y - axis will be same as earlier one or 2.30x10⁻⁹ N . Component along x axis will cancel out.
c ) Electric field in case 1 and case 2 will be
= 2.30x10⁻⁹ / 1 x 10⁻⁹
= 2.3 N / C , because field is force per unit charge. The sane field will be in case 2 .
Answer:
Find the attachments sequence wise for complete solution.
