(a) Briefly explain what is meant by threshold effect in FM receivers. Why is this effect not present in PM receivers?

Create a program named PaintingDemo that instantiates an array of eight Room objects and demonstrates the Room methods. The Room

Serggg [28]

Answer:

Explanation:

Code used will be like

using System;

using System.Collections.Generic;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

namespace PaintingWall

{

class Room

{

public int length, width, height,Area,Gallons;

public Room(int l,int w,int h)

{

length = l;

width = w;

height = h;

}

private int getLength()

{

return length;

}

private int getWidth()

{

return width;

}

private int getHeight()

{

return height;

}

public void WallAreaAndNumberGallons()

{

Area = getLength() * getHeight() * getWidth();

if (Area < 350)

{

Gallons = 1;

}

else if (Area > 350)

{

Gallons = 2;

}

Console.WriteLine ("The area of the Room is " + Area);

Console.WriteLine("The number of gallons paint needed to paint the Room is " + Gallons);

}

class PaintingDemo

{

static void Main(string[] args)

{

int l, w, h;

Room[] r = new Room[8];

for (int i = 0; i <= 7; i++)

{

Console.WriteLine("Room "+(i+1));

Console.Write("Enter Length : ");

l = Convert.ToInt32(Console.ReadLine() );

Console.Write("Enter Width : ");

w = Convert.ToInt32(Console.ReadLine());

Console.Write("Enter Height : ");

h= Convert.ToInt32(Console.ReadLine());

r[i] = new Room(l,w,h);

Console.WriteLine();

}

for (int i = 0; i <= 7; i++)

{

Console.WriteLine("Room " + (i + 1));

r[i].WallAreaAndNumberGallons();

}

Console.ReadKey();

}

3 0

3 years ago

A new 1.2-ha suburban residential development is to be drained by a storm sewer that connects to the municipal drainage system.

valentinak56 [21]

Answer:

Time of concentration, $t_d$ ⇒ 1280 min

Peak runoff rate, $Q_p$ ⇒4.185 ff³/s

Explanation:

See detailed explanation

6 0

3 years ago

A box-shaped aquarium has horizontal dimensions 0.5 m by 1 m, and depth 0.5 m, and is filled two-thirds of the way to the surfac

antoniya [11.8K]

Answer:

3270 N/m^2

Explanation:

we can calculate the pressure difference between the bottom and surface of the tank by applying the equation for the net vertical pressure

Py = - Ph ( g ± a )

for a downward movement

Py = - Ph ( g - a ) ------ ( 1 )

From the above data given will be

p = 1000 kg/m^3, h = 2/3 * 0.5 = 0.33 m , a =2g , g = 9.81

input values into equation 1 becomes

Py = -Ph ( g - 2g ) = Phg ------ ( 3 )

Py = 1000 * 0.33 * 9.81

= 3270 N/m^2

6 0

3 years ago

A heat pump with an ideal compressor operates between 0.2 MPa and 1 MPa. Refrigerant R134a flows through the system at a rate of

solmaris [256]

Answer:

The mass flow rate of refrigerant is 0.352 kg/s

Explanation:

Considering the cycle of an ideal heat pump, provided in the attachment, we first find enthalpy at state B and D. For that purpose, we use property tables of refrigerant R134a:

At State A:

From table, we see the enthalpy and entropy value of saturated vapor at 0.2 MPa. Therefore:

ha = 244.5 KJ/kg

Sa = 0.93788 KJ/kg.k

At State B:

Since, the process from state A to B is isentropic. Therefore,

Sb = Sa = 0.93788 KJ/Kg

From table, we see the enthalpy value of super heated vapor at 1 MPa and Sb. Therefore:

hb = 256.85 KJ/kg (By interpolation)

At State C:

From table, we see the enthalpy and entropy value of saturated liquid at 1 MPa. Therefore:

hc = 107.34 KJ/kg

Now, from the diagram it is very clear that:

Heat Loss = m(hb = hc)

m = (Heat Loss)/(hb - hc)

where,

m = mass flow rate = ?

Heat Loss = (180,000 Btu/hr)(1.05506 KJ/1 Btu)(1 hr/3600 sec)

Heat Loss = 52.753 KW

Therefore,

m = (52.753 KJ/s)/(256.85 KJ/kg - 107.34 KJ/kg)

m = 0.352 kg/s

5 0

3 years ago

Ultimate tensile strength is: (a) The stress at 0.2% strain (b) The stress at the onset of plastic deformation (c) The stress at

MissTica

Answer:

By definition the ultimate tensile strength is the maximum stress in the stress-strain deformation. The stress at 0.2% strain, the stress at the onset of plastic deformation, the stress at the end of the elastic deformation and the stress at the fracture correspond, by definition, to other points of the stress-strain curve.

Explanation:

4 0

3 years ago