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maria [59]
3 years ago
5

3. Each bangle printed by a 3D printer has a mass of exactly 25 g of metal. If the density of the metal is 14 g/cm3, what length

of a wire 1 mm in radius is needed to produce each bangle? Find your answer to the tenths place.
Mathematics
1 answer:
nadezda [96]3 years ago
5 0

Answer:

57.0 cm

Step-by-step explanation:

We know that the radius of the filament is 1 mm. Taken to cm, is 0.1 cm.

From the density and mass, we can get the volumen of the bangle:

d = m/V ----> V = m/d

Solving for V:

V = 25 / 14 = 1.79 cm^3

We can assume that a bangle is a cylinder so:

V = π*r^2*h

1.79 = 3.14 * 0.1^2 * h

h = 1.79 / 3.14 * 0.01

h = 57.0 cm

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\left[\begin{array}{ccc}140\\160\\200\end{array}\right]

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\bf M=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]

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P_2=[P].[M]^2=[P].[M].[M]

M^2=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]\times \left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]

     =\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]

\therefore\;\;\;\;\;\;\;\;\;\;\;P_2=\left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right] \times\left[\begin{array}{ccc}130\\300\\70\end{array}\right] = \left[\begin{array}{ccc}140\\160\\200\end{array}\right]

b) - With in migration process, 500 people are numbered. There will be after a long time,

After\;inifinite\;period=[M]^n.[P]

Then,\;we\;get\;the\;same\;result\;if\;we\;measure [M]^n=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right]

                                   =\left[\begin{array}{ccc}140\\160\\200\end{array}\right]

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------
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