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3 years ago

The cube measures 3.0-ft on all sides and has a density of 3.1 slugs/ft3. How much does it weigh?

Engineering

1 answer:

kodGreya [7K]3 years ago

4 0

Answer:

W = 2695.14 lb

Explanation:

given,

side of cube = 3 ft

density of the cube = 3.1 slugs/ft³

we know,

$density = \dfrac{mass}{volume}$

mass = density x volume

volume = 3³ = 27 ft³

mass = 3.1 x 27

m = 83.7 slugs.

weight calculation

converting mass from slug to pound

weight of 1 slug is equal to 32.2 lb

now,

weight of the cube is equal to

W = 83.7 slugs x 32.2 lb/slug

W = 2695.14 lb

hence, weight is equal to W = 2695.14 lb

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Answer:

A) The sketches for the required planes were drawn in the first attachment [1 2 1] and the second attachment [1 2 -4].

B) The closest distance between planes are d₁₂₁=a/√6 and d₁₂₋₄=a/√21 with lattice constant a.

C) Five posible directions that electrons can move on the surface of a [1 0 0] silicon crystal are: |0 0 1|, |0 1 3|, |0 1 1|, |0 3 1| and |0 0 1|.

Compleated question:

1. Miller Indices:

a. Sketch (on separate plots) the (121) and (12-4) planes for a face centered cubic crystal structure.

b. What are the closest distances between planes (called d₁₂₁ and d₁₂₋₄)?

c. List five possible directions (using the Miller Indices) the electron can move on the surface of a (100) silicon crystal.

Explanation:

A)To draw a plane in a face centered cubic lattice, you have to follow these instructions:

1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment) and the planes has 3 main coeficients shown as [l m n]

2- The coordinates of that plane are written as: π:[1/a₀ 1/b₀ 1/c₀] (if one of the coordinates is 0, for example [1 1 0], c₀ is ∞, therefore that plane never cross the direction c).

3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.

4- Join the points.

In this case, for [1 2 1]:

$l=1=1/a_0 \rightarrow a_0=1$

$m=2=2/b_0 \rightarrow b_0=0.5$

$n=1=1/c_0 \rightarrow c_0=1$

for $[1 2 \overline{4}]$ :

$l=1=1/a_0 \rightarrow a_0=1$

$m=2=2/b_0 \rightarrow b_0=0.5$

$n=\overline{4}=-4/c_0 \rightarrow c_0=-0.25$

B) The closest distance between planes with the same Miller indices can be calculated as:

With $\pi:[l m n]$ ,the distance is $d_{lmn}= \displaystyle \frac{a}{\sqrt{l^2+m^2+n^2}}$ with lattice constant a.

In this case, for [1 2 1]:

$d_{121}= \displaystyle \frac{a}{\sqrt{1^2+2^2+1^2}}=\frac{a}{\sqrt{6}}=0.41a$

for $[1 2 \overline{4}]$ :

$d_{12\overline{4}}= \displaystyle \frac{a}{\sqrt{1^2+2^2+(-4)^2}}=\frac{a}{\sqrt{21}}=0.22a$

C) The possible directions that electrons can move on a surface of a crystallographic plane are the directions contain in that plane that point in the direction between nuclei. In a silicon crystal, an fcc structure, in the plane [1 0 0], we can point in the directions between the nuclei in the vertex (0 0 0) and e nuclei in each other vertex. Also, we can point in the direction between the nuclei in the vertex (0 0 0) and e nuclei in the center of the face of the adjacent crystals above and sideways. Therefore:

dir₁=|0 0 1|

dir₂=|0 0.5 1.5|≡|0 1 3|

dir₃=|0 1 1|

dir₄=|0 1.5 0.5|≡|0 3 1|

dir₅=|0 0 1|

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