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;
}
}
complete question
A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?
Answer:
3.03 V 0.184 W
2.499 mV 125*10^-9 W
Explanation:
First, apply voltage-divider principle to the input circuit: 1
*5
= 4.545 V
The voltage produced by the voltage-controlled source is:
A_voc*V_i = 4.545 V
We can find voltage across the load, again by using voltage-divider principle:
V_o = A_voc*V_i*(R_o/R_l+R_o)
= 4.545*(100/100+50)
= 3.03 V
Now we can determine delivered power:
P_L = V_o^2/R_L
= 0.184 W
Apply voltage-divider principle to the circuit:
V_o = (R_o/R_o+R_s)*V_s
= 50/50+100*10^3*5
= 2.499 mV
Now we can determine delivered power:
P_l = V_o^2/R_l
= 125*10^-9 W
Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.
Answer:
Consider A is square matrix of order 4 x 4 generated using magic function. Augmented matrix can be generated using:
Aug=[A eye(size(A))]
Above command is tested in MATLAB command window and is attached in figure below
Answer:
2.0%
Explanation:
Percentage of aggregate = 94%
Specific gravity = 2.65
Specific gravity of asphalt = 1.9
Density of mix = 147pcf = 147lb/ft³
Total weight of mix: (volume = 1ft³)
= (147lb/ft³)(1ft³)
= 147lb
Percentage weight of asphalt in<u> mix:</u>
100% - 94%
= 6%
Weight of asphalt binders
= 6% x 147lb
= 8.82lb
Weight of aggregate in mix:
= 94% x 147
= 138.18lb
Specific weight of asphalt binder:
(Gab)(Yw)
Yw = specific Weight of water
= 62.4lb
Gab = specific gravity of asphalt binder
= 1.0
(62.4lb)(1.0)
= 62.4 lb/ft³
Volume of asphalt in binder:
8.82/62.4
= 0.14ft³
Specific weight of binder in mix:
2.65 x 62.4lb/ft³
= 165.36 lb/ft³
Volume of aggregate:
= 138.18/165.36
= 0.84ft³
Volume of void in the mix:
1ft³ - 0.84ft³ - 0.14ft³
= 0.02ft³
<u>The percentage of void in total mix:</u>
VTM = (0.02ft³/1ft³)100
= 2.0%