Answer:
None of these
Explanation:
There are different types of amplifiers, and each has different characteristics.
- Voltage amplifier needs high input and low output resistance.
- Current amplifier needs Low Input and High Output resistance.
- Trans-conductance amplifier Low Input and High Output resistance.
- Trans-Resistance amplifier requires High Input and Low output resistance.
Therefore, the correct answer is "None of these "
Answer:
Explanation:
Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:
![Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]](https://tex.z-dn.net/?f=Q_%7Bwater%7D%20%3D%20%281.4%5C%2CL%29%5Ccdot%28%5Cfrac%7B1%5C%2Cm%5E%7B3%7D%7D%7B1000%5C%2CL%7D%20%29%5Ccdot%20%281000%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%20%29%5Ccdot%20%5B%284.187%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20%5E%7B%5Ctextdegree%7DC%7D%20%29%5Ccdot%20%28100%5E%7B%5Ctextdegree%7DC-25%5E%7B%5Ctextdegree%7DC%29%2B2257%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D%5D)
The heat liberated by the LP gas is:
A kilogram of LP gas has a minimum combustion power of 
. Then, the required mass is:
Answer:
False ( B )
Explanation:
considering that the wind turbine is a horizontal axis turbine
Power generated/extracted by the turbine can be calculated as
P = n * 1/2 *<em> p</em> *Av^3
where: n = turbine efficiency
<em>p = air density </em>
<em> </em>A = πd^2 / 4
v = speed
From the above equation it can seen that increasing the Blade radius by 10% will increase the Blade Area which will in turn increase the value of the power extracted by the wind turbine
Answer:
import java.util.*;
public class Main {
public static void main(String[] args) {
double milesPerGallon = 0;
int totalMiles = 0;
int totalGallons = 0;
double totalMPG = 0;
Scanner input = new Scanner(System.in);
while(true){
System.out.print("Enter the miles driven: ");
int miles = input.nextInt();
if(miles <= 0)
break;
else{
System.out.print("Enter the gallons used: ");
int gallons = input.nextInt();
totalMiles += miles;
totalGallons += gallons;
milesPerGallon = (double) miles/gallons;
totalMPG = (double) totalMiles / totalGallons;
System.out.printf("Miles per gallon for this trip is: %.1f\n", milesPerGallon);
System.out.printf("Total miles per gallon is: %.1f\n", totalMPG);
}
}
}
}
Explanation:
Initialize the variables
Create a while loop that iterates until the specified condition is met inside the loop
Inside the loop, ask the user to enter the miles. If the miles is less than or equal to 0, stop the loop. Otherwise, for each trip do the following: Ask the user to enter the gallons. Add the miles and gallons to totalMiles and totalGallons respectively. Calculate the milesPerGallon (divide miles by gallons). Calculate the totalMPG (divide totalMiles by totalGallons). Print the miles per gallon and total miles per gallon.
Answer:
(4.5125 * 10^-3 kg.m^2)ω_A^2
Explanation:
solution:
Moments of inertia:
I = mk^2
Gear A: I_A = (1)(0.030 m)^2 = 0.9*10^-3 kg.m^2
Gear B: I_B = (4)(0.075 m)^2 = 22.5*10^-3 kg.m^2
Gear C: I_C = (9)(0.100 m)^2 = 90*10^-3 kg.m^2
Let r_A be the radius of gear A, r_1 the outer radius of gear B, r_2 the inner radius of gear B, and r_C the radius of gear C.
r_A=50 mm
r_1 =100 mm
r_2 =50 mm
r_C=150 mm
At the contact point between gears A and B,
r_1*ω_b = r_A*ω_A
ω_b = r_A/r_1*ω_A
= 0.5ω_A
At the contact point between gear B and C.
At the contact point between gears A and B,
r_C*ω_C = r_2*ω_B
ω_C = r_2/r_C*ω_B
= 0.1667ω_A
kinetic energy T = 1/2*I_A*ω_A^2+1/2*I_B*ω_B^2+1/2*I_C*ω_C^2
=(4.5125 * 10^-3 kg.m^2)ω_A^2
