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Sonbull [250]
4 years ago
13

Answer true or false 3.Individual people decide what will be produced in a command oconomy

Engineering
2 answers:
Pie4 years ago
7 0

Answer:

False

Explanation:

The government decides the productions.

Leto [7]4 years ago
3 0
False

In a command economy, the government controls major aspects of economic production. The government decides the means of production and owns the industries that produce goods and services for the public. The government prices and produces goods and services that it thinks benefits the people.
You might be interested in
A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is appli
lukranit [14]

Answer:

For the Top Side

- Strain ε  = 0.00021739

- Elongation is 0.00260868 cm

For The Right side

- Strain ε  = 0.00021739

-Elongation is 0.00347826 cm

Explanation:

Given the data in the question;

Length of the squared titanium plate = 12 cm by 12 cm = 0.12 m by 0.12 m

Thickness = 5 mm = 0.005 m

Force to the Top F_t = 15 kN = 15000 Newton

Force to the right F_r = 20 kN = 20000 Newton

elastic modulus, E = 115 GPa = 115 × 10⁹ pascal

Now, For the Top Side;

- Strain = σ/E = F_t  / AE

we substitute

= 15000 / ( 0.12 × 0.005 × (115 × 10⁹) )

= 15000 / 69000000

Strain ε  = 0.00021739

- Elongation

Δl = ε × l

we substitute

Δl = 0.00021739 ×  12 cm

Δl = 0.00260868 cm

Hence, Elongation is 0.00260868 cm

For The Right side

- Strain = σ/E = F_r  / AE

we substitute

Strain = 20000 / ( 0.12 × 0.005 × (115 × 10⁹) )

= 20000 / 69000000

Strain ε = 0.000289855

- Elongation

Δl = ε × l

we substitute

Δl = 0.000289855×  12 cm

Δl = 0.00347826 cm

Hence, Elongation is 0.00347826 cm

5 0
3 years ago
Consider liquid n-hexane in a 50-mm diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the
ra1l [238]

The evaporation rate of the n-Hexane is 7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}

<u>Explanation</u>:

This is a situation regarding diffusing A through non-diffusing B.

A = n-Hexane B=Air

Where the molar flux is provided by,

N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}

\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}

P_{t}=1 a t m=101325 P a\\

\text { so, } P_{A 1}= the vapor pressure at hexane 25 \mathrm{C} =20158.2 \mathrm{Pa}

For wind, assume negligible hexane is present, hence P_{A 2}=0

Now,

\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}

\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}

P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\

=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}

=90873.57 \mathrm{Pa}

R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}

z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\

where T = 298 K

substituting all in the equation, we get

\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}

=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\

Now,Flux \times area  = Molar rate of evaporation

Evaporation rate = 0.004 \mathrm{mol} / \mathrm{m}^{2}-5 \mathrm{x}\left(\pi \mathrm{d}^{2} / 4 \mathrm{m}^{2}\right)=0.004 \times(3.14 \times 0.05 \times 0.05 / 4)

Evaporation rate =7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}

6 0
3 years ago
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
Isormophous phase diagram
shusha [124]

Answer:

Phase diagrams represent the relationship between temperature and the composition of phases present at equilibrium. An isomorphous system is one in which the solid has the same structure for all compositions. The phase diagram shown is the diagram for Cu-Ni, which is an isomorphous alloy system.

Hope it help you friend

6 0
3 years ago
A common structural feature of membrane lipids is their amphipathic nature. For example, in phosphatidylcholine, the two fatty a
BabaBlast [244]

Answer:

A. Phosphatidylethanolamine:

phosphoethanolamine <em>(hydrophilic)</em> 2 fatty acids <em>(hydrophobic )</em>

<em />

B. Sphingomyelin:

phosphocholine <em>(hydrophilic)</em> ceramide (sphingosine + 1 fatty acid) <em>(hydrophobic )</em>

<em />

C. Galactosylcerebroside:

D-galactose <em>(hydrophilic)</em> ceramide <em>(hydrophobic )</em>

<em />

D. Ganglioside:

oligosaccharide <em>(hydrophilic)</em> ceramide <em>(hydrophobic )</em>

<em />

E. Cholesterol:

-OH group <em>(hydrophilic)</em> hydrocarbon ring skeleton <em>(hydrophobic )</em>

<em />

Explanation:

The amphipathic nature of lipids shows  that all lipids have a <em>hydrophilic </em>(water-loving or polar end) and a <em>hydrophobic</em><em>  </em>(water fearing or non-polar end.)

<em />

The various lipids in question, and their ampiphatic groups are detailed above.

6 0
3 years ago
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