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2 years ago

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Consider an airfoil in a wind tunnel (i.e., a wing that spans the entire test section). Prove that the lift per unit span can be

obtained from the pressure distributions on the top and bottom walls of the wind tunnel (i.e., from the pressure distributions on the walls above and below the airfoil). For this problem, assume that the pressure distributions on the upper and lower walls of the wind tunnel are pu(x) and pl(x), respectively. They are functions of the x-coordinates. You will need to apply the momentum equation in integral form in the y-direction.

1 answer:

Brrunno [24]2 years ago

7 0

Answer:

Step-by-step explanation:

to solve this proving kindly check the attached files

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Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

3 0

2 years ago

David wants to install a fence around his garden. The garden is rectangular and has an area of 240 square feet. One side of the

oee [108]

Answer:

$996

Step-by-step explanation:

The rectangular plot has an area that is the product of its length and width. We are given the width as 12 feet, and the area as 240 ft², so we can find the length from ...

... A = L×W

... 240 ft² = L×(12 ft)

... 240 ft²/(12 ft) = L = 20 ft

Opposite sides of the rectangle are the same length, so the cost of fence for a side of a given length will be the sum of the costs of the opposites sides.

The 12 ft side has one segment that is $18 per foot, and one that is $15 per foot. For the 20 ft sides, both are $15 per foot. Then the total cost can be figured from ...

... (12 ft)·($18/ft + $15/ft) + (20 ft)·($15/ft +$15/ft) = 12·$33 +20·$30 = $996

5 0

2 years ago

Carla packed this box with 1 centimeter cubes what is the volume of the box

shusha [124]

Answer: $60\ cm^3$

Step-by-step explanation:

<h3> The complete exercise is attached.</h3>

You can observe in the picture attached that the box is a rectangular prism.

The volume of a rectangular prism can be found with this formula:

$V=lwh$

Where "l" is the length, "w" is the width and "h" is the height.

You know that the lenght of each side of those cubes is 1 centimeter. Therefore, you can multiply the number of cubes on each side of the box by 1 centimeter in order to find the lenght, the width and the height of the box:

$l=(4)(1\ cm)=4\ cm\\\\w=(3)(1\ cm)=3\ cm\\\\h=(5)(1\ cm)=5\ cm$

Now you can substitute the lenght, the width and the height of the box into the formula shown at the beginning of the explanation:

$V=(4\ cm)(3\ cm)(5\ cm)$

Finally, evaluating, you get that the volume of the box is:

$V=60\ cm^3$

4 0

2 years ago

Please help me answer this ASAP! The two-way table shows the number of ninth and tenth graders who prefer going to sporting even

Elis [28]

<h2>Answer</h2>

0.43

<h2>Explanation</h2>

Remember that $RelativeFrecuency=\frac{Frecuency}{SumOfAllFrecuencies}$

Since the problem is telling us "Among tenth graders", we must focus on the 10th graders row only. From the row, we can infer that the frequency is the number of 10th graders who prefer going to sporting events, so $Frecuency=6$ . Now, the sum of all frequencies will be the sum of all the 10th graders, so $SumOfAllFrecuencies=6+8=14$ . Let's replace the values:

$RelativeFrecuency=\frac{Frecuency}{SumOfAllFrecuencies}$

$RelativeFrecuency=\frac{6}{14}$

$RelativeFrecuency=0.4285$

And rounded to the nearest hundredth:

$RelativeFrecuency=0.43$

4 0

2 years ago

The percent of licensed U.S. drivers (from a recent year) that are female is 48.60. Of the females, 5.03% are age 19 and under;

docker41 [41]

Answer:

a) 51.4

b) answer attached

c) 48.59% female

Step-by-step explanation:

a) Male driver = 100-48.6 = 51.4%

b) answer attached below

c) probability that out of 20-64 group a randomly selected sample is female

( 39.54 ÷ 81.36 ) x 100

= 48.59% chance of her being a female

6 0

2 years ago