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Serga [27]
3 years ago
7

a triangle is defined by the three vertices. write the following functions of the triangle class assume that the point class has

already been written for you you may add helper functions as needed
Engineering
1 answer:
Fofino [41]3 years ago
3 0

Answer:

The triangle() function is an inbuilt function in p5.js which is used to draw a triangle in a plane. This function accepts three vertices of triangle.

Syntax:

triangle(x1, y1, x2, y2, x3, y3)

Below programs illustrates the triangle() function in p5.js:

function setup() {  

   createCanvas(400, 400);  

}  

 

function draw() {  

   background(220);  

   fill('lightgreen');  

 

   // A triangle at (100, 250), (250, 170) and (330, 300)    

   triangle(100, 250, 250, 170, 330, 300);  

}  

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A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pip
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Answer:

Part A: (d/D=0.1)

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Part B:(d/D=0.01)

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Explanation:

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DeltaV=\frac{\pi * DeltaP}{8*u*l}(R^{4}-r^{4} -\frac{(R^{2}-r^{2})}{ln\frac{R}{r}}^{2})

Above equation can be written as:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{d}{D} )^{4}+\frac{(1-(\frac{d}{D})^{2})}{ln\frac{d}{D}}^{2})

First Consider no wire i.e d/D=0

Above expression will become:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0)^{4}+\frac{(1-(0)^{2})}{ln0}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}

Part A: (d/D=0.1)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.1)^{4}+\frac{(1-(0.1)^{2})}{ln0.1}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.574

DeltaV percent=\frac{(\frac{\pi*R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.574}{\frac{\pi*R^{4}*DeltaP}{8*u*l} }*100

DeltaV percent=\frac{1-0.574}{1}*100

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.783

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DeltaV percent=\frac{1-0.783}{1}*100

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