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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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Answer:

The original length of the specimen l_{o} = 104.7 mm

Explanation:

Original diameter d_{o} = 30 mm

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Final length l_{1} = 105.20 mm

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We know that the relation between the shear modulus & elastic modulus is given by

G = \frac{E}{2(1 + \mu)}

25.5 = \frac{65.5}{2 (1 + \mu)}

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We know that the possion's ratio is given by

\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }

{\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}

{\frac{change \ in \ length}{l_{o} } = 0.00476

\frac{l_{1} - l_{o}  }{l_{o}  } = 0.00476

\frac{l_{1} }{l_{o} } = 1.00476

Final length l_{o} = 105.2 m

Original length

l_{o} = \frac{105.2}{1.00476}

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Both of the technicians are correct.

<u>Explanation:</u>

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It is also used to determine the direction the front wheels are pointing. While the scrub radius is the intersecting point of the vertical center line of front tires with the imaginary line drawn from the steering knuckles.

Thus both the technicians are saying correct. The thrust angle and scrub radius are used to determine the alignment of wheels, if there is any misalignment in wheels then it needs to make it correct to prevent accidents.

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Len [333]

Answer:

Some of the internal strain energy is relieved.

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