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Dima020 [189]
2 years ago
14

The accompanying specific gravity values describe various wood types used in construction. 0.320.350.360.360.370.380.400.400.40

0.410.410.420.420.420.420.420.430.44 0.450.460.460.470.480.480.490.510.54 0.540.550.580.630.660.660.670.680.78 Construct a stem-and-leaf display using repeated stems. (Enter numbers from smallest to largest separated by spaces. Enter NONE for stems with no values.)
Engineering
1 answer:
marysya [2.9K]2 years ago
8 0

Answer:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\  \ \\ {0.7} & {\vert} & {8} \ \ \end{array}

Explanation:

Given

0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,

0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,

0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,

0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.

Required

Plot a steam and leaf display for the given data

Start by categorizing the data by their tenth values:

0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.

0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,

0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.

0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.

0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.

0.78.

The 0.3's is will be plotted as thus:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}

The 0.4's is as follows:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \ \end{array}

The 0.5's is as follows:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}

The 0.6's is as thus:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}

Lastly, the 0.7's is as thus:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ {0.7} & {\vert} & {8} \ \ \end{array}

The combined steam and leaf plot is:

\begin{array}{ccc}{Steam} & {\vert} & {Leaf}  \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\  \ \\ {0.7} & {\vert} & {8} \ \ \end{array}

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A). When t= 3 s, the acceleration is

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dv/dt = 1- 2(3)

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B. How far it traveled in 3 sec

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Distance = 27/2 - 9

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Consider a NACA 2412 airfoil in a low-speed flow at zero degrees angle of attack and a Reynolds number of 8.9·106 . Calculate th
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Answer:

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b) pressure drag is 20%

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Ans) Given,

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Also, Cd = Cdf + Cdp

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Now, for zero degree angle of attack for NACA 2412, Cd = 0.0075

Also, Cd = Cdf + Cdp

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Hence, pressure drag is 20% of pressure drag due to flow seperation  Ans) Given,

NACA 2412 airfoil

Re = 8.9 x 10^6

We know, for turbulent flow ,drag coefficient, Cdf = 0.074 / Re^0.2

=> Cdf = 0.074 / (8.9 x 10^6)^0.2

=> Cdf = 0.003

For both side of plate, Cd = 2 x 0.003 = 0.006

For zero degree angle of attack for NACA 2412, Cdf = 0.006

Also, Cd = Cdf + Cdp

=> 0.006 = 0.006 + Cdp

=> Cdp = 0

Hence, pressure drag is zero

Now, for zero degree angle of attack for NACA 2412, Cd = 0.0075

Also, Cd = Cdf + Cdp

=> 0.0075 = 0.006 + Cdp

=> Cdp = 0.0075 - 0.006

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Hence, Pressure drag percentage = (Cdp / Cdf) x 100

=> Pressure drag percent = (0.0015/0.0075) x 100 = 20 %

Hence, pressure drag is 20% of pressure drag due to flow seperation  

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And the output screenshot is: [Attached]

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