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2 years ago

You can divide a surface by drawing a line through it

Engineering

2 answers:

tatuchka [14]2 years ago

8 0

Answer:

yes you can

Explanation:

hope this helps, have a good day :-)

SpyIntel [72]2 years ago

6 0

Answer:

Explanation:

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wo companies, Ajax Co. and Boho Inc., were negotiating a merger. In the course of the negotiations, an Ajax representative told

boyakko [2]

Answer:

Yes

Explanation:

If the Ajax representative fails to correct the previous statement this can cause misrepresentation.

4 0

2 years ago

A specific internal combustion engine has a displacement volume VD of 5.6 liters. The processes within each cylinder of the engi

Kisachek [45]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0

3 years ago

A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an

Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = $1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}$

Cyclone is to 80 % efficient

So particulate remaining = $0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136$

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %

4 0

3 years ago

For this question you must write a java class called Rectangle and a client class called RectangleClient. The partial Rectangle

Alex Ar [27]

Answer:

Java program is given below. You can get .class after you execute java programs, You can attach those files along with .java classes given , Those .class files are generated ones.

Explanation:

//Rectangle.java class

public class Rectangle {

private int x;

private int y;

private int width;

private int height;

// constructs a new Rectangle with the given x,y, width, and height

public Rectangle(int x, int y, int w, int h)

{

this.x=x;

this.y=y;

this.width=w;

this.height=h;

}

// returns the fields' values

public int getX()

{

return x;

}

public int getY()

{

return y;

}

public int getWidth()

{

return width;

}

public int getHeight()

{

return height;

}

// returns a string such as “Coordinate is (5,12) and dimension is 4x8” where 4 is width and 8 is height. public String toString()

public String toString()

{

String str="";

//add x coordidate , y-coordinate , width, height and area to str and return

str+="Coordinate is ("+x+","+y+")";

str+=" and dimension is : "+width+"x"+height;

str+=" Area is "+(width*height);

return str;

}

public void changeSize(int w,int h)

{

width=w;

height=h;

}

======================

//main.java

class Main {

public static void main(String[] args) {

//System.out.println("Hello world!");

//create an object of class Rectangle

Rectangle rect=new Rectangle(5,12,4,8);

//print info of rect using toString method

System.out.println(rect.toString());

//chamge width and height

rect.changeSize(3,10);

//print info of rect using toString method

System.out.println(rect.toString());

}

==========================================================================================

//Output

Coordinate is (5,12) and dimension is : 4x8 Area is 32

Coordinate is (5,12) and dimension is : 3x10 Area is 30

========================================================================================

6 0

3 years ago

Race car is accelerating and has a velocity of 10 m/s @ t=0. It completes a lap on a circular track of 400 m in 14 seconds. Calc

wariber [46]

Answer:

component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²

magnitude of acceleration is 22.98 m/s²

Explanation:

given data

velocity = 10 m/s

initial time to = 0

distance s = 400 m

time t = 14 s

to find out

components and magnitude of acceleration after the car has travelled 200 m

solution

first we find the radius of circular track that is

we know distance S = 2πR

400 = 2πR

R = 63.66 m

and tangential acceleration is

S = ut + 0.5 ×at²

here u is initial speed and t is time and S is distance

400 = 10 × 14 + 0.5 ×a (14)²

a = 3.37 m/s²

and here tangential acceleration is constant

so velocity at distance 200 m

v² - u² = 2 a S

v² = 10² + 2 ( 3.37) 200

v = 38.05 m/s

so radial acceleration at distance 200 m

ar = $\frac{v^2}{R}$

ar = $\frac{38.05^2}{63.66}$

ar = 22.74 m/s²

so magnitude of total acceleration is

A = $\sqrt{a^2 + ar^2}$

A = $\sqrt{3.37^2 + 22.74^2}$

A = 22.98 m/s²

so magnitude of acceleration is 22.98 m/s²

8 0

3 years ago