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Maurinko [17]
3 years ago
15

I don’t understand this

Engineering
1 answer:
blondinia [14]3 years ago
7 0

Answer:

Sorry for the delayed response- Right now I don't have time to give you the answer, but I really want to help so I'll try to phrase it in a easier way to understand things: Basically what you need to do for this problem is find the area of the base of the figure (which means length x width) and then you would simply find the volume of 160cm^{2} by finding the length of each side of the figure, find the length of the figure, find the height of the figure and then find the radius.

Have an amazing day and I hope this can somewhat help :)

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A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.6
amm1812

1) a_x=4.287+2.772x\\a_y=-5.579+2.772y

2) 8.418

Explanation:

1)

The two components of the velocity field in x and y for the field in this problem are:

u=1.85+2.05x+0.656y

v=0.754-2.18x-2.05y

The x-component and y-component of the acceleration field can be found using the following equations:

a_x=\frac{du}{dt}+u\frac{du}{dx}+v\frac{du}{dy}

a_y=\frac{dv}{dt}+u\frac{dv}{dx}+v\frac{dv}{dy}

The derivatives in this problem are:

\frac{du}{dt}=0

\frac{dv}{dt}=0

\frac{du}{dx}=2.05

\frac{du}{dy}=0.656

\frac{dv}{dx}=-2.18

\frac{dv}{dy}=-2.05

Substituting, we find:

a_x=0+(1.85+2.05x+0.656y)(2.05)+(0.754-2.18x-2.05y)(0.656)=\\a_x=4.287+2.772x

And

a_y=0+(1.85+2.05x+0.656y)(-2.18)+(0.754-2.18x-2.05y)(-2.05)=\\a_y=-5.579+2.772y

2)

In this part of the problem, we want to find the acceleration at the point

(x,y) = (-1,5)

So we have

x = -1

y = 5

First of all, we substitute these values of x and y into the expression for the components of the acceleration field:

a_x=4.287+2.772x\\a_y=-5.579+2.772y

And so we find:

a_x=4.287+2.772(-1)=1.515\\a_y=-5.579+2.772(5)=8.281

And finally, we find the magnitude of the acceleration simply by applying Pythagorean's theorem:

a=\sqrt{a_x^2+a_y^2}=\sqrt{1.515^2+8.281^2}=8.418

4 0
3 years ago
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