Multiple Choice

Determine the mass density of an oil if 0.3 tonnes of the oil occupies a volume of 4m.

Yuki888 [10]

Answer:

Do you mean 4m^3 and 3.0 tones?

Explanation:

solution:

Mass = m = 3.0 tones

- 1 ton = 1,000 kg

= 3.0 × 1,000

= 3,000 kg

volume = v = 4m^3

Required:

Mass density of oil = p = ?

We know that;

$p = \frac{mass}{volume} = \frac{m}{v} = \frac{3000}{4} = 750kg |m^{3} ans$

The answer is:

750kg / m^3

8 0

3 years ago

Cost and benefit, environmental effects, ethical or legal implications, health and safety considerations, and sustainability are

kykrilka [37]

Answer:

the major items for engineers to consider during the design process.

7 0

3 years ago

Silicon carbide nanowires of diameter D = 15 nm can be grown onto a solid siliconcarbide surface by carefully depositing droplet

Elina [12.6K]

Answer:

For unfined package = 1.30 × 10^-3 W.

Dor fined package = 8.64 × 10^-3 W.

Explanation:

STEP ONE: The first thing to do is to determine the cross sectional area and input the value into the convectional resistance formula/equation in order to determine the heat rate for the unfined package.

Thus, Area = (dimension of the electronic device) ^2. = (10 × 10^-6)^2 m.

Convectional Resistance = (100 × 10^-9)÷ (10 × 10^-6)^2 × 490 = 2.04 K/W.

Heat transfer coefficient = 1/ (10 × 10^-7)^2 × 10^5. = 1 × 10^5 K/W.

Thus, the heat rate for the unfined package = 2{(85 - 20) ÷ (10^5 + 2.04)} = 1.30 × 10^-3 W.

STEP TWO: Determine the surface area of each fin and the prime area.

Surface area of each fin = {300 + 15/4} × 10^-9 × 10^-9 × π × 15 = 1.43 × 10^-14 m^2.

Prime area = (10^-6 × 10)^2 - (200 × 200) × π × [ ( 15 × 10^-9)^2 ÷ 4] = 9.29 × 10^-11 m^2.

STEP THREE: Determine the efficiency

G ={ [300 + 15/4) × 10^9 } × [ 4 × 10^5÷ 490 × 15 × 10^-9] = 7.09 × 10^-2.

Efficiency= tanh ( 7.09 × 10^-2) ÷ 7.09 × 10^-2 = 0.998.

Thus, the coefficient = 1 - (1.43 × 10^-14 × (200 × 200) ÷ (6.65 × 10^-10) × (1 - 0.998) = 0.99.

STEP FOUR: Determine the heat rate for the finned package.

Thermal resistance = 1/ (10^5 × 0.999 × 10^-10 × 6.65 = 1.50 × 10^4 K/W.

The heat rate for the finned package = 2 { 85 - 20} ÷ { 1.50 × 10^4 + 2.04} = 8.64 × 10^-3 W

4 0

3 years ago

Write a program to calculate the area of circle, rectangle using a class. The program would prompt a user to select (1) circle o

cricket20 [7]

Answer:

The following program written in c++:

#include <iostream>

using namespace std;

class Circle //define class

{

private : //access modifier

double pi, radius, area; // set double type variables

public :

Circle (double radius = 0.0) //define constructor

{

pi = 3.1415;

this->radius = radius;

}

void setRadius (double radius) // define void type method

{

if (radius < 0){ // if the radius is negative don't take it

cout << "Radius can't be negative." << endl;

return;

}

this->radius = radius;

}

double getArea () // define double type method

{

area = pi * radius * radius;

return area;

}

};

class Rectangle //define class

{

private: //access modifier

double length ,breadth ,area; // set double type private data types

public:

Rectangle (double length = 0.0, double breadth = 0.0) // define constructor

{

this->length = length;

this->breadth = breadth;

}

void setDimension (double length , double breadth ) // define void type method

{

if (length < 0 || breadth < 0){ //set if condition, if the dimensions is negative than don't take it

cout << "Dimensions can't be negative." << endl;

return;

}

this->length = length;

this->breadth = breadth;

}

double getArea () // define double type method

{

area = length * breadth;

return area;

}

};

int main() //main function

{

int choice; //integer type variable

double radius, length, breadth;

Circle c;

Rectangle r;

cout << "Select ( 1 ) Circle or select ( 2 ) Rectangle : "; // get the choice of user

cin >> choice;

switch (choice)

{

case 1 : cout << "Enter radius of circle : ";

cin >> radius;

c.setRadius(radius);

cout << "Area of circle is : " << c.getArea();

break;

case 2 : cout << "Enter dimensions of rectangle : ";

cin >> length >> breadth;

r.setDimension(length,breadth);

cout << "Area of rectangle is : " << r.getArea();

break;

default : cout << "Invalid Selection...";

}

return 0;

}

Explanation:

Here, we define two classes "Circle" and "Rectangle"

Then, we create first two function which get the value from the user.

Then, we create the second two functions which return the calculation of the following formulas.

3 0

3 years ago

The stagnation chamber of a wind tunnel is connected to a high-pressure airbottle farm which is outside the laboratory building.

Natasha2012 [34]

This question is not complete, the complete question is;

The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005

Answer:

the length of the pipe is 11583 in or 965.25 ft

Explanation:

Given the data in the question;

Static pressure ratio; p1/p2 = 10

friction coefficient f = 0.005

diameter of pipe, D =4 inch

first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so

4fL $_{max}$ / D = 57.915

we substitute

(4×0.005×L $_{max}$ ) / 4 = 57.915

0.005L $_{max}$ = 57.915

L $_{max}$ = 57.915 / 0.005

L $_{max}$ = 11583 in

Therefore, the length of the pipe is 11583 in or 965.25 ft

6 0

3 years ago