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lana [24]
3 years ago
15

In the frequency domain, you view a signal whose highest frequency is fH=4.3kHz, and lowest frequency is fL=300Hz. What is the s

ignal’s frequency bandwidth?
Engineering
1 answer:
Tema [17]3 years ago
7 0

Answer:

BW=\Delta f= f_H-f_L=4300Hz-300Hz=4000Hz=4kHz

Explanation:

Bandwidth is the difference between the upper and lower frequencies in a continuous set of frequencies. The bandwidth may be determined by use of the following formula:

BW=f_H-f_L\\\\Where:\\\\f_H=Upper\hspace{3}cutoff\hspace{3}frequency=4.3kHz=4300Hz\\f_L=Lower\hspace{3}cutoff\hspace{3}frequency=300Hz

Hence the signal’s frequency bandwidth is:

BW=4300-300=4000Hz=4kHz

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Type the correct answer in the box. Spell all words correctly.
Semmy [17]

Answer:

Anne is a mechanical engineer.

Explanation:

There are many different types of engineers <em>(mechanical, industrial, electrical, chemical and civil) </em>around the world.<em> </em>When a job involves <em>designing machines</em>, it falls under <u>mechanical engineering</u>. Such job requires a person to be as creative as possible, so he/she can come up with an innovative design that will be reasonable when used.

Designing machines includes<em> spacecraft</em>s. Therefore, Anne is a mechanical engineer working for NASA.

3 0
4 years ago
Create a program named IntegerFacts whose Main() method declares an array of 10 integers.Call a method named FillArray to intera
musickatia [10]

Answer:

C# codee

using System;

class IntegerFacts

{

static void Main()

{

int[] x = new int[10];

int sum = 0;

int siz = 0, high = 0, low = 0, avg = 0;

siz = FillArray(x);

Statistics(x, ref high, ref low, ref sum, ref avg, siz);

Console.Write("The highest value is " + high);

Console.Write("\nThe lowest value is " + low);

Console.Write("\nThe sum of the values is " + sum);

Console.Write("\nThe average is " + avg);

}

static void Statistics (int [] b, ref int h, ref int l, ref int s, ref int a, int size)

{

if (size == 0)

h = l = s = a = 0;

else

{

int i =0;

l = 999;

h = 0;

s = 0;

for (; i < size;++i)

{

if(b[i] > h)

h = b[i];

if(b[i] < l)

l = b[i];

s += b[i];

}

a = s / size;

}

}

static int FillArray (int[] a)

{

int i = 0, count = 0;

int intTemp =0 ;

string temp;

for(i = 0 ; i < 10 ; i++)

{

Console.Write("Enter element number"+(i+1) +" : ");

temp = Console.ReadLine();

if (int.TryParse(temp, out intTemp)) {

if (intTemp != 999)

{

a[i] = intTemp;

count++;

}

else

break;

}

else

{

Console.Write("\n\nOops.. You entered a wrong number. Try again \n\n");

i--;

}

}

return count;

}

}

Explanation:

The output of the above program is given in the attached file.

8 0
4 years ago
The end of a cylindrical liquid cryogenic propellant tank in free space is to be protected from external (solar) radiation by pl
prohojiy [21]

If the temperature of the shield is 338 kelvin. Then the heat flux through the tank will be 25.3 Watt per square meter.

<h3>What is heat flux?</h3>

The increase in heat energy movement through a particular surface is known as heat flux, and the heat flux density is the absolute temperature per unit area.

Assume the view factor between the tank and the shield is unity; all surfaces are diffuse and gray, and the surroundings are at 0 K.

It is given that T= 100 K, ε₁ = ε₂ = 0. 10, \varepsilon_t = 0.20, and GS= 1250 W/m².

Then we have

The temperature of the shield will be

\rm \alpha _sG_s - \varepsilon _1 E_b (T_s) - \dot{q}_{ST} = 0 ...1

and

\rm q''_{12}=\dfrac{ \sigma (T_{1}^{4} - T_{2}^{2})}{\frac{1}{\varepsilon _1 }+ \frac{1}{\varepsilon _2} -1}} ...2

Then from equations 1 and 2, we have

\rm \alpha _sG_s - \varepsilon _1 E_b (T_s) - \dfrac{ \sigma (T_{1}^{4} - T_{2}^{2})}{\frac{1}{\varepsilon _1 }+ \frac{1}{\varepsilon _2} -1}} = 0

Then the value of \rm T_s will be

\rm T_s =\left [ \dfrac{\alpha _sGs+\left ( \dfrac{\sigma T_1^4}{\frac{1}{\varepsilon _1}+\frac{1}{\varepsilon _2} - 1} \right )}{\sigma \left ( \varepsilon _1 + \dfrac{1}{\frac{1}{\varepsilon _1} + \frac{1}{\varepsilon _2}-1} \right )} \right ] ^{\dfrac{1}{4}}

Put all the values, then we have

\rm T_s = \left [ \dfrac{0.05 \times + \left ( \dfrac{\sigma (100)^4}{\frac{1}{0.1}+\frac{1}{0.05}-1} \right )}{\sigma \left ( 0.05 + \dfrac{1}{\frac{1}{0.1}+\frac{1}{0.05} - 1} \right )} \right ]^{\dfrac{1}{4}} \\\\\\T_s = 338 \ K

Then the heat flux will be

\rm q"_{ST}=\dfrac{\sigma (T_S^4 - T_t^4)}{\frac{1}{\varepsilon _1} + \frac{1}{\varepsilon _2} - 1} \\\\\\q"_{ST}=\dfrac{5.67 \times 10^{-8}(388^4-100^4)}{\frac{1}{0.1}+\frac{1}{0.05}-1}\\\\\\q"_{ST} = 25.3 \ W/m^2

More about the heat flux link is given below.

brainly.com/question/12913016

#SPJ1

8 0
3 years ago
What is an isochoric process? b) Can heat be exchanged in an isochoric process? c) A 100L container holding an ideal gas at an i
oksano4ka [1.4K]

Answer:

a)A constant volume process is called isochoric process.

b)Yes

c)Work =0

Explanation:

Isochoric process:

 A constant volume process is called isochoric process.

In constant volume process work done on the system or work done by the system will remain zero .Because we know that work done give as

work = PΔV

Where P is pressure and ΔV is the change in volume.

For constant volume process ΔV = 0⇒ Work =0

Yes heat transfer can be take place in isochoric process.Because we know that temperature difference leads to transfer of heat.

Given that

Initial P=10 MPa

Final pressure =15 MPa

Volume = 100 L

Here volume of gas is constant so the work work done will be zero.

4 0
4 years ago
Write a java program which will produce the following output. No variables are needed but the program must use numbers and math
Sav [38]

The question did not include the output to be produced. However, below are the outputs.

My age is 21

Twice my age is 42

Three times my age is 63

ANSWER:

public class NewClasss{

   public static void main(String[] args) {

   int x,y,z;

   x = 21;

   y = 2*x;

   z=3*x;

   System.out.println("My age is " + x);

   System.out.println("twice my age is " + y);

   System.out.println("three times my age is " + z);

   

   }

  }

8 0
3 years ago
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