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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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<h3>How to code?</h3>
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Consider a resistor made of pure silicon with a cross-sectional area pf 0.5 μm2, and a length of 50 μm. What is the resistance o
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Answer: 24 pA

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As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.

Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵  Ω  cm.

The resistance R of a given resistor, is expressed by the following formula:

R = ρ L / A

Replacing by the values for resistivity, L and A, we have

R = 2.1. 10⁵ Ω  cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2

R = 2.1. 10¹¹ Ω

Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:

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Explanation:

Solution

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The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

Diameter of the tube, D = 1cm =0.01 m

Electrical heat rate, q =76 W/m

Wall Temperature, Ts = 370 K

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From the properties table of engine oil we can deduce as follows:

thermal conductivity, k =0.139 W/m .K

Density, ρ = 854 kg/m³

Specific heat, cp = 2120 J/kg.K

(a) Thus

The wall heat flux is given as follows:

qs = q/πD

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= 2419.16 W/m²

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The oil mean temperature is given as follows:

Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)

Tb =370 - 11/24 * (2419.16 * 0.005/0.139)

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(b) The center line temperature is given below:

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The The axial gradient of the mean temperature is given below:

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dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120

dTb/dx = 19.81 K/m

(d) The heat transfer coefficient is given below:

h =48/11 (k/D)

=48/11 (0.139/0.01)

h =60.65 W/m². K

8 0
3 years ago
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