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just olya [345]

4 years ago

7

Digital information can be used to modulate an analog carrier wave. Like analog modulation, the carrier’s amplitude, frequency,

and phase angle can be used to represent logical data ("1s" and "0s").
True

False

2 answers:

Fudgin [204]4 years ago

6 0

Answer:

1) True, The carrier’s amplitude, frequency, and phase angle can be used to represent logical data (“1s” and “0s”).

Explanation:

Dovator [93]4 years ago

4 0

the answer is true..hope this helps :))

You might be interested in

This program will store roster and rating information for a soccer team. Coaches rate players during tryouts to ensure a balance

kari74 [83]

Answer:

Using C++ language

#include <iostream>

#include <vector>

using namespace std;

int main()

{

//definition

vector<int> jerseyNumber;

vector<int> rating;

int temp;

//loop

for (int i = 1; i <= 5; i++)

{

cout << "Enter player " << i

<< "'s jersey number: ";

cin >> temp;

jerseyNumber.push_back(temp);

cout << "Enter player " << i

<< "'s rating: ";

cin >> temp;

rating.push_back(temp);

cout << endl;

}

cout << "ROSTER" << endl;

for (int i = 0; i < 5; i++)

cout << "Player " << i + 1 << " -- "

<< "Jersey number: " << jerseyNumber.at(i)

<< ", Rating: " << rating.at(i) << endl;

char option;

'

while (true)

{

cout << "MENU" << endl;

cout << "a - Add player" << endl;

cout << "d - Remove player" << endl;

cout << "u - Update player rating" << endl;

cout << "r - Output players above a rating"

<< endl;

cout << "o - Output roster" << endl;

cout << "q - Quit" << endl << endl;

cout << "Choose an option: ";

cin >> option;

switch (option)

{

case 'a':

case 'A':

cout << "Enter a new player's"

<< "jersey number: ";

cin >> temp;

jerseyNumber.push_back(temp);

cout << "Enter the player's rating: ";

cin >> temp;

rating.push_back(temp);

break;

case 'd':

case 'D':

cout << "Enter a jersey number: ";

cin >> temp;

int i;

for (i = 0; i < jerseyNumber.size();

i++)

{

if (jerseyNumber.at(i) == temp)

{

jerseyNumber.erase(

jerseyNumber.begin() + i);

rating.erase(rating.begin() + i);

break;

}

}

break;

case 'u':

case 'U':

cout << "Enter a jersey number: ";

cin >> temp;

for (int i = 0; i < jerseyNumber.size();

i++)

{

if (jerseyNumber.at(i) == temp)

{

cout << "Enter a new rating "

<< "for player: ";

cin >> temp;

rating.at(i) = temp;

break;

}

}

break;

case 'r':

case 'R':

cout << "Enter a rating: ";

cin >> temp;

cout << "\nABOVE " << temp << endl;

for (int i = 0; i < jerseyNumber.size();

i++)

if (rating.at(i) > temp)

cout << "Player " << i + 1

<< " -- "

<< "Jersey number: "

<< jerseyNumber.at(i)

<< ", Rating: "

<< rating.at(i) << endl;

break;

case 'o':

case 'O':

cout << "ROSTER" << endl;

for (int i = 0; i < jerseyNumber.size();

i++)

cout << "Player " << i + 1 << " -- "

<< "Jersey number: "

<< jerseyNumber.at(i) << ", Rating: "

<< rating.at(i) << endl;

break;

case 'q':

return 0;

default: cout << "Invalid menu option."

<< " Try again." << endl;

}

}

}

6 0

3 years ago

Air at 293k and 1atm flow over a flat plate at 5m/s. The plate is 5m wide and 6m long. (a) Determine the boundary layer thicknes

loris [4]

Answer:

a). 8.67 x $10^{-3}$ m

b).0.3011 m

c).0.0719 m

d).0.2137 N

e).1.792 N

Explanation:

Given :

Temperature of air, T = 293 K

Air Velocity, U = 5 m/s

Length of the plate is L = 6 m

Width of the plate is b = 5 m

Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X $10^{-5}$ Pa-s

We know density of air is ρ = 1.21 kg / $m^{3}$

Now we can find the Reyonld no at x = 1 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 1}{1.822\times 10^{-5} }$

Re = 332052.6

Therefore the flow is laminar.

Hence boundary layer thickness is

δ = $\frac{5.x}{\sqrt{Re}}$

= $\frac{5\times 1}{\sqrt{332052.6}}$

= 8.67 x $10^{-3}$ m

a). Boundary layer thickness at x = 1 is δ = 8.67 X $10^{-3}$ m

b). Given Re = 100000

Therefore the critical distance from the leading edge can be found by,

Re = $\frac{\rho .U.x}{\mu }$

100000 = $\frac{1.21\times5\times x}{1.822 \times10^{-5}}$

x = 0.3011 m

c). Given x = 3 m from the leading edge

The Reyonld no at x = 3 m from the leading edge

Re = $\frac{\rho .U.x}{\mu }$

Re = $\frac{1.21 \times 5\times 3}{1.822\times 10^{-5} }$

Re = 996158.06

Therefore the flow is turbulent.

Therefore for a turbulent flow, boundary layer thickness is

δ = $\frac{0.38\times x}{Re^{\frac{1}{5}}}$

= $\frac{0.38\times 3}{996158.06^{\frac{1}{5}}}$

= 0.0719 m

d). Distance from the leading edge upto which the flow will be laminar,

Re = $\frac{\rho \times U\times x}{\mu }$

5 X $10^{5}$ = $\frac{1.21 \times 5\times x}{1.822\times 10^{-5}}}$

x = 1.505 m

We know that the force acting on the plate is

$F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

and $C_{D}$ at x= 1.505 for a laminar flow is = $\frac{1.328}{\sqrt{Re}}$

= $\frac{1.328}{\sqrt{5\times10 ^{5}}}$

= 1.878 x $10^{-3}$

Therefore, $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho \times A\times U^{2}$

= $\frac{1}{2}\times 1.878\times 10^{-3}\times 1.21\times (5\times 1.505)\times 5^{2}$

= 0.2137 N

e). The flow is turbulent at the end of the plate.

Re = $\frac{\rho \times U\times x}{\mu }$

= $\frac{1.21 \times 5\times 6}{1.822\times 10^{-5} }$

= 1992316

Therefore $C_{D}$ = $\frac{0.072}{Re^{\frac{1}{5}}}$

= $\frac{0.072}{1992316^{\frac{1}{5}}}$

= 3.95 x $10^{-3}$

Therefore $F_{D}$ = $\frac{1}{2}\times C_{D}\times \rho\times A\times U^{2}$

= $\frac{1}{2}\times 3.95\times 10^{-3}\times 1.21\times (5\times 6)\times 5^{2}$

= 1.792 N

3 0

3 years ago

Imagine the arc of a football as it flies through the air. How does this motion illustrate classical mechanics?

BigorU [14]

Answer: Maybe A

Explanation:

7 0

1 year ago

Read 2 more answers

If a material is found to be in the tertiary phase of creep, the following procedure should be implemented:

kirill [66]

When a material is found to be in the tertiary phase of creep, the following procedure should be implemented that is the component should be replaced immediately. Therefore, Option C is correct.

<h3>What do you mean by a tertiary degree of creep?</h3>

Tertiary Creep has an extended creep rate and terminates when the material breaks or ruptures. It is related to each necking and formation of grain boundary voids. The wide variety of possible stress-temperature- time combos is infinite.

Therefore, When a material is found to be in the tertiary phase of creep, the following procedure should be implemented that is the component should be replaced immediately. Option C is correct.

Learn more about creep:

brainly.com/question/10565749

#SPJ1

8 0

2 years ago

Water flows in a tube that has a diameter of D= 0.1 m. Determine the Reynolds number if the average velocity is 10 diameters per

Cloud [144]

Answer:

a) $Re_{D} = 111896.745$ , b) $Re_{D} = 1.119\times 10^{-7}$

Explanation:

a) The Reynolds number for the water flowing in a circular tube is:

$Re_{D} = \frac{\rho\cdot v\cdot D}{\mu}$

Let assume that density and dynamic viscosity at 25 °C are $997\,\frac{kg}{m^{3}}$ $0.891\times 10^{-3}\,\frac{kg}{m\cdot s}$ , respectively. Then:

$Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m}{s} )\cdot (0.1\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }$

$Re_{D} = 111896.745$

b) The result is:

$Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (10^{-6}\,\frac{m}{s} )\cdot (10^{-7}\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }$

$Re_{D} = 1.119\times 10^{-7}$

6 0

4 years ago