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olchik [2.2K]
3 years ago
6

In a short essay, discuss the question, "How are you an innovator?"

Engineering
1 answer:
iragen [17]3 years ago
7 0

Answer:

Being innovative means doing things differently or doing things that have never been done before. An innovator is someone who has embraced this idea and creates environments in which employees are given the tools and resources to challenge the status quo, push boundaries and achieve growth.

Explanation:

Hope it helps..

But it's a little bit long..

Correct me if I'm wrong..

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
Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i
Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\

Work done by pump

W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W

∴ Power input to the pump 20.7088 kW

6 0
3 years ago
A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3
labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass co_2 = 1 kg

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of co_2 = 44

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

V_1 = \frac{m RT_1}{P_1}

       =\frac{1*0.189*298}{104}

V_1 = 0.5415 m3

V_2 = \frac{m RT_2}{P_2}

     =\frac{1*0.189*584}{1068}

V_1 = 0.1033 m3

WORK DONE

W =P_{avg}*{V_2-V_1}

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

\Delta U  = m *c_v*{V_2-V_1}

\Delta U  = 1*0.657*(584-298)

\Delta U  =187.902 kJ

HEAT TRANSFER

Q = \Delta U  +W

   = 187.902 +(-256.46)

Q = -68.859 kJ

7 0
3 years ago
How might an operations manager alter operations to meet customer demand? Name at least two ways.
Citrus2011 [14]
One way is manager changes itself and the other one is the same thing i think.
4 0
3 years ago
For each of the resistors shown below, use Ohm's law to calculate the unknown quantity, Be sure to put your answer in proper eng
daser333 [38]

Answer:

the hurts my brain sorry bud cant help

Explanation:

6 0
3 years ago
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