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3 years ago

A___ remote control can be an advantage to an

Engineering

2 answers:

KATRIN_1 [288]3 years ago

8 0

Answer:

Wireless

Explanation:

Elina [12.6K]3 years ago

8 0

I think the correct choice would me wireless

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State the number of terms for each of the following algebraic expression 2x+1

harina [27]

Answer:

Expressions are made up of terms.

A term is a product of factors.

Coefficient is the numerical factor in the term

Before moving to terms like monomials, binomials, and polynomials, like and unlike terms are discussed.

When terms have the same algebraic factors, they are like terms.

When terms have different algebraic factors, they are unlike terms.

Explanation:

Hi please follow me also if you can and thanks.

6 0

3 years ago

Which term defines the amount of mechanical work an engine can do per unit of heat energy it uses?

skad [1K]

Answer:

Explanation:

is the because that's the amount of work in making machine can do producing heat

7 0

3 years ago

Applying the Entropy Balance: Closed Systems Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston–

Mrrafil [7]

Answer:

a) the amount of energy produced in kJ/K is 0.73145 kJ/K

b) the amount of energy produced in kJ/K is 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

Explanation:

Draw the T-s diagram.

$C_p$ = 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280

R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar

Δs = $m[c_p ln(\frac{T_2}{T_1}) - Rln(\frac{P_2}{P_1})]$

Substitute all parameters in the equation

Δs = $5[(0.939) ln(\frac{520}{280}) - (\frac{8.314}{44.01})ln(\frac{20}{2})]$

Δs = 5 kg × 0.14629 kJ/kg.K

= 0.73145 kJ/K

Δs = $m[\frac{s^0(T_2) - s^0(T_1)}{M} - Rln(\frac{P_2}{P_1})]$

Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K

T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K

R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar

Δs = $5[\frac{236.575 - 211.376}{44.01} - (\frac{8.314}{44.01})ln(\frac{20}{2})]$

= 5 kg (0.13795 kJ/kg.K)

= 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

7 0

3 years ago

Write the implementation (.cpp file) of the Player class from the previous exercise. Again, the class contains:

hoa [83]

Answer:

//Define the header file

#ifndef PLAYER_H

#define PLAYER_H

//header file.

#include <string>

//Use the standard namespace.

using namespace std;

//Define the class Player.

class Player

{

//Declare the required data members.

string name;

int score;

public:

//Declare the required

//member functions.

void setName(string par_name);

void setScore(int par_score);

string getName();

int getScore();

}

//End the definition

//of the header file.

#endif

Player.cpp:

//Include the "Player.h" header file,

#include "Player.h"

//Define the setName() function.

void Player::setName(string par_name)

{

name = par_name;

}

//Define the setScore() function.

void Player::setScore(int par_score)

{

score = par_score;

}

//Define the getName() function.

string Player::getName()

{

return name;

}

//Define the getScore() function.

int Player::getScore()

{

return score;

}

7 0

3 years ago

For beam design, the cross section dimensions of the beam are determined: A. Based on allowable normal stress, but the allowable

noname [10]

Answer:

B. Based on the allowable shear stress, but the allowable normal stress should always be checked to be sure it is not exceeded

Explanation:

Shear stress is analyzed to determine the shear forces along the lenght of the beam. This is represented in a shear force diagram. The beam cross sectional design is determined in such a way as to minimize the shear stress. Allowable normal stress should always be checked in a structure if failure is to be prevented.

8 0

3 years ago