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Dafna1 [17]
3 years ago
14

You live on a street that runs East to West. You just had 2 inche of snow and you live on the North side of the street. You retu

rn from class at 2PM and notice all the snow on your sidewalk is gone but across the street it is still there. No one removed the snow. How did it go away?
Engineering
1 answer:
RSB [31]3 years ago
5 0

Answer:

The heat from the sun melted it

Explanation:

If the street runs east to west, houses on the south (across the street) will project shadows on their sidewalk, while the northern sidewalk will be illuminated. This is for the northern hemisphere, on the southern hemisphere it would be the other way around.

You might be interested in
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
Which branch of engineering studies the physical behavior of metallic elements?
olga55 [171]

Material engineering studies the physical behavior of metallic elements.

Answer: Option C

<u>Explanation: </u>

Material Engineering is the creation and learning about the materials at an atomic level. An engineers from this branch focus on material and model its characteristics using the computer.

Also, they combine the knowledge of solid-states, metallurgy, chemistry and ceramics to the application level. It also has a great role in building the future with the advancing study in nanotechnology, biotechnology, etc. Simply, these are meant to have vivid applications in future life.

6 0
3 years ago
Read 2 more answers
A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o
victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width= 25 mm = 25\times 10^{-3} m

thickness = 6.5 mm = 6.5\times 10^{-3} m

crack length 2c = 0.5 mm at centre of specimen

\sigma _{applied} =  1000 N/cross sectional area

stress intensity factor  =  k  will be

\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}

                   = 6.154\times 10^{6} Pa

we know that

k =\sigma_{applied} (\sqrt{\pi C})

  =6.154\sqrt{\pi (2.5\times 10^{-04})}          [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

ifK_C = 1.15 Mpa m^{1/2} then load will be

Kc = \sigma _{frac}(\sqrt{\pi C})

1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}

\sigma _{frac} = 41.04 MPa

load = \sigma _{frac}\times Area

load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N

LAOD = 6669.86 N

3 0
3 years ago
What forces are not present in space
ss7ja [257]

Answer:

C.) Weight and distance I believe

Explanation:

7 0
3 years ago
Because assembly language is so close in nature to machine language, it is referred to as a ____________.
My name is Ann [436]

Answer:

symbolic machine code.

Explanation:

The instructions in the language are closely linked to the machine's architecture.

5 0
4 years ago
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