Answer:
R=1923Ω
Explanation:
Resistivity(R) of copper wire at 20 degrees Celsius is 1.72x10^-8Ωm.
Coil length(L) of the wire=37.0m
Cross-sectional area of the conductor or wire (A) = πr^2
A= π * (2.053/1000)/2=3.31*10^-6
To calculate for the resistance (R):
R=ρ*L/A
R=(1.72*10^8)*(37.0)/(3.31*10^-6)
R=1922.65Ω
Approximately, R=1923Ω
Answer:
The mass flow rate of the mixture in the manifold is 6.654 kg/min
Explanation;
In this question, we are asked to calculate mass flow rate of the mixture in the manifold
Please check attachment for complete solution and step by step explanation.
Answer:
Explanation: Here it is: 67 Hope that helps! :)
Answer:
// Program is written in Java Programming Language
// Comments are used for explanatory purpose
import java.util.*;
public class FlipCoin
{
public static void main(String[] args)
{
// Declare Scanner
Scanner input = new Scanner (System.in);
int flips;
// Prompt to enter number of toss or flips
System.out.print("Number of Flips: ");
flips = input.nextInt();
if (flips > 0)
{
HeadsOrTails();
}
}
}
public static String HeadsOrTails(Random rand)
{
// Simulate the coin tosses.
for (int count = 0; count < flips; count++)
{
rand = new Random();
if (rand.nextInt(2) == 0) {
System.out.println("Tails"); }
else {
System.out.println("Heads"); }
rand = 0;
}
}
