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;
}
}
Answer:
98°C
Explanation:
Total surface area of cylindrical fin = πr² + 2πrl , r = 0.015m; l= 0.1m; π =22/7
22/7*(0.015)² + 22/7*0.015*0.1 = 7.07 X 10∧-4 + 47.1 X 10∧-4 = (54.17 X 10∧-4)m²
Temperature change, t = (50 - 25)°C = 25°C = 298K
Hence, Temperature = 150 X (54.17 X 10∧-4) X 298/123 = 242.14/124 = 2.00K =
∴ Temperature change = 2.00K
But temperature, T= (373 - 2)K = 371 K
In °C = (371 - 273)K = 98°C
Answer:
Amount of air left in the cylinder=m
=0.357 Kg
The amount of heat transfer=Q=0
Explanation:
Given
Initial pressure=P1=300 KPa
Initial volume=V1=0.2
Initial temperature=T
=20 C
Final Volume=
=0.1
Using gas equation
m1==(300*0.2)/(.287*293)
m1=0.714 Kg
Similarly
m2=(P2*V2)/R*T2
m2=(300*0.1)/(0.287*293)
m2=0.357 Kg
Now calculate mass of air left,where me is the mass of air left.
me=m2-m1
me=0.715-0.357
mass of air left=me=0.357 Kg
To find heat transfer we need to apply energy balance equation.
Where me=m1-m2
And as the temperature remains constant,hence the enthalpy also remains constant.
h1=h2=he=h
Q=(me-(m1-m2))*h
me=m1-me
Thus heat transfer=Q=0
Answer:
If i am correct It should be 1/4 of an inch
Explanation:
Sorry but i can't quite explain
