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ratelena [41]
3 years ago
14

The major resisting force in gravity dam? ​

Engineering
2 answers:
Sloan [31]3 years ago
8 0

Explanation:

Self weight of dam.............

inna [77]3 years ago
6 0

The self weight of dam.

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Amanda and Tyler opened a business that specializes in shipping liquids, such as milk, juice, and water, in cylindrical containe
USPshnik [31]

Answer:

circleType.h

#ifndef circleType_H

#define circleType_H

class circleType

{

public:

void print();

void setRadius(double r);

//Function to set the radius.

//Postcondition: if (r >= 0) radius = r;

// otherwise radius = 0;

double getRadius();

//Function to return the radius.

//Postcondition: The value of radius is returned.

double area();

//Function to return the area of a circle.

//Postcondition: Area is calculated and returned.

double circumference();

//Function to return the circumference of a circle.

//Postcondition: Circumference is calculated and returned.

circleType(double r = 0);

//Constructor with a default parameter.

//Radius is set according to the parameter.

//The default value of the radius is 0.0;

//Postcondition: radius = r;

private:

double radius;

};

#endif

circleTypeImpl.cpp

#include <iostream>

#include "circleType.h"

using namespace std;

void circleType::print()

{

cout << "Radius = " << radius

<< ", area = " << area()

<< ", circumference = " << circumference();

}

void circleType::setRadius(double r)

{

if (r >= 0)

radius = r;

else

radius = 0;

}

double circleType::getRadius()

{

return radius;

}

double circleType::area()

{

return 3.1416 * radius * radius;

}

double circleType::circumference()

{

return 2 * 3.1416 * radius;

}

circleType::circleType(double r)

{

setRadius(r);

}

cylinderType.h

#ifndef cylinderType_H

#define cylinderType_H

#include "circleType.h"

class cylinderType: public circleType

{

public:

void print();

void setHeight(double);

double getHeight();

double volume();

double area();

//returns surface area

cylinderType(double = 0, double = 0);

private:

double height;

};

#endif

cylinderTypeImpl.cpp

#include <iostream>

#include "circleType.h"

#include "cylinderType.h"

using namespace std;

cylinderType::cylinderType(double r, double h)

: circleType(r)

{

setHeight(h);

}

void cylinderType::print()

{

cout << "Radius = " << getRadius()

<< ", height = " << height

<< ", surface area = " << area()

<< ", volume = " << volume();

}

void cylinderType::setHeight(double h)

{

if (h >= 0)

height = h;

else

height = 0;

}

double cylinderType::getHeight()

{

return height;

}

double cylinderType::area()

{

return 2 * 3.1416 * getRadius() * (getRadius() + height);

}

double cylinderType::volume()

{

return 3.1416 * getRadius() * getRadius() * height;

}

main.cpp

#include <iostream>

#include <iomanip>

using namespace std;

#include "cylinderType.h"

int main()

{

double radius,height;

double shippingCostPerLi,paintCost,shippingCost=0.0;

 

cout << fixed << showpoint;

cout << setprecision(2);

cout<<"Enter the radius :";

cin>>radius;

 

cout<<"Enter the Height of the cylinder :";

cin>>height;

 

 

cout<<"Enter the shipping cost per liter :$";

cin>>shippingCostPerLi;

 

 

//Creating an instance of CylinderType by passing the radius and height as arguments

cylinderType ct(radius,height);

 

double surfaceArea=ct.area();

double vol=ct.volume();

 

 

shippingCost+=vol*28.32*shippingCostPerLi;

 

char ch;

 

cout<<"Do you want the paint the container (y/n)?";

cin>>ch;

if(ch=='y' || ch=='Y')

{

cout<<"Enter the paint cost per sq foot :$";

cin>>paintCost;    

shippingCost+=surfaceArea*paintCost;    

}    

cout<<"Total Shipping Cost :$"<<shippingCost<<endl;

 

return 0;

}

3 0
3 years ago
A 1-kW electric resistance heater submerged in 10-kg water is turned on and kept on for 15 min. During the process, 400 kJ of he
hichkok12 [17]

Answer:

ΔT=  11.94 °C

Explanation:

Given that

mass of water = 10 kh

Time t= 15 min

Heat lot from water = 400  KJ

Heat input to the water = 1  KW

Heat input the water= 1 x 15 x 60

                                =900 KJ

By heat balancing

Heat supply - heat rejected = Heat gain by water

As we know that heat capacity of water

C_p=4.187 \frac{KJ}{kg-K}

Q=mC_p\Delta T

Now by putting the values

900 - 400 = 10 x 4.187 x ΔT

So  rise in temperature of water ΔT=  11.94 °C

6 0
3 years ago
Suppose that a class CalendarDate has been defined for storing a calendar date with month, day and year components. (In our sect
laiz [17]

Answer:

note:

find the attachment

7 0
3 years ago
If extension done in a stork position, during backward bending in a standing position of a special test for lumbar spine, produc
Klio2033 [76]
The answer is 7 because I just took the test!
5 0
2 years ago
A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the
Tanya [424]

Answer:

a) 23.89 < -25.84 Ω

b) 31.38 < 25.84 A

c) 0.9323 leading

Explanation:

A) Calculate the load Impedance

current on load side = 0.75 p.u

power factor angle = 25.84

I_{load} = 0.75 < 25.84°

attached below is the remaining part of the solution

<u>B) Find the input current on the primary side in real units </u>

load current in primary = 31.38 < 25.84 A

<u>C) find the input power factor </u>

power factor = 0.9323 leading

<em></em>

<em>attached below is the detailed solution </em>

8 0
3 years ago
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