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sukhopar [10]
3 years ago
8

What are the main causes of injuries when using forklifts?

Engineering
1 answer:
vagabundo [1.1K]3 years ago
5 0

The forklift overturning is a very common way of getting injured from a forklift. Overturning the forklift means it tips over onto it's side due to the operator turning it too fast.

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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 commercial refrigerator with refrigerant -134a as the working fluid is used to keep the refrigerated space at -30C by rejectin
Mariana [72]

Answer:

a) 0.487

b) refrigeration load = 5.46w

c) cop = 2.24

d)ref load max = 12.43kw

Explanation:

6 0
3 years ago
Read 3 more answers
Given below are the measured streamflows in cfs from a storm of 6-hour duration on a stream having a drainage area of 185 mi^2.
sertanlavr [38]

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

<u>Derive the unit hydrograph using the inverse procedure </u>

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh *  duration

        = 29700 * 6 hours ( 216000 secs )

        = 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000  / 185 mi^2

                    = 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh /  volume of runoff in equivalent depth

                                = 50 ft^3 / 1.49 in  =  33.56 ft^3/sec.in

5 0
3 years ago
In what type of automobile is a transaxle most commonly found?
user100 [1]

Answer: vehicles with a front engine and FWD or a rear engine and RWD.

Explanation But the transaxle can also be integrated into the rear axle on cars with a front engine and rear-wheel drive. The transaxle is in the rear where the differential would be rather than beside the engine.

7 0
2 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Assoli18 [71]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}  

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

8 0
3 years ago
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