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3 years ago

Technologies that allow for instant worldwide communication include

Engineering

2 answers:

hram777 [196]3 years ago

7 0

Answer:

mobile phones and internet access

Explanation:

I got it right on my quiz

Gennadij [26K]3 years ago

6 0

Answer: B

Explanation:

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96/64 reduced to its lowest term

Marina CMI [18]

Answer:

3/2

Explanation:

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3 years ago

Acertain foundation will experience a bearing capacity failurewhen it is subjected to a downward load of 2200 kN. Using ASD with

ehidna [41]

Answer:

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Explanation:

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3 years ago

Explain briefly the power analogy between the fluid “water” and the Word of God.

melomori [17]

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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

A plate (A-C) is connected to steelflat bars by pinsat A and B. Member A-E consists of two 6mm by 25mm parallel flat bars. At C,

juin [17]

Answer:

stress_ac = 5.333 MPa

shear stress_c = 1.763 MPa

Explanation:

Given:

- The missing figure is in the attachment.

- The dimensions of member AC = ( 6 x 25 ) mm x 2

- The diameter of the pin d = 19 mm

- Load at point A is P = 2 kN

Find:

- Find the axial stress in AE and the shear stress in pin C.

Solution:

- The stress in member AE can be calculated using component of force P along the member AE as follows:

stress_ac = P*cos(Q) / A_ae

Where, Angle Q: A_E_B and A_ac: cross sectional area of member AE.

cos(Q) = 4 / 5 ..... From figure ( trigonometry )

A_ae = 0.006*0.025*2 = 3*10^-4 m^2

Hence,

stress_ae = 2*(4/5) / 3*10^-4

stress_ae = 5.333 MPa

- The force at pin C can be evaluated by taking moments about C equal zero:

(M)_c = P*6 - F_eb*3

0 = P*6 - F_eb*3

F_eb = 0.5*P

- Sum of horizontal forces for member AC is zero:

P - F_eb - F_c = 0

F_c = 0.5*P

- The shear stress of double shear bolt is given by an expression:

shear stress = shear force / 2*A_pin

Where, The area of the pin C is:

A_pin = pi*d^2 / 4

A_pin = pi*0.019^2 / 4 = 2.8353*10^-4 m^2

Hence,

shear stress = 0.5*P / 2*A_pin

shear stress = 0.5*2 / 2*2.8353*10^-4

shear stress = 1.763 MPa

7 0

3 years ago