Concept explainers
(a)
To sketch: The plane within the cubic cell for
(b)
To sketch: The plane within the cubic cell for
(c)
To sketch: The plane within the cubic cell for
(d)
To sketch: The plane within the cubic cell for
(e)
To sketch:The plane within the cubic cell for
(f)
To sketch:The plane within the cubic cell for
(g)
To sketch: The plane within the cubic cell for
(h)
To sketch:The plane within the cubic cell for
(i)
To sketch: The plane within the cubic cell for
(j)
To sketch:The plane within the cubic cell for
(k)
To sketch: The plane within the cubic cell for
(l)
To sketch:The plane within the cubic cell for
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MindTap Engineering for Askeland/Wright's The Science and Engineering of Materials, 7th Edition, [Instant Access], 2 terms (12 months)
- 2. Draw the directions in the cubic unit cell. A:[00 T]. B:[120]. C:[I11]. D:[21 T]. 3. Draw the planes in the cubic unit cell. A:(111). B:(030). C: (102).arrow_forwardDraw the following directions within the empty unit cell below [101], [113], and [212].arrow_forwardSketch the following planes and directions within a cubic unit cell: Describe the procedures for both of the direction and the plane. (a) [110] (b) [221] (c) [410] (d) [012] (e) [321] (f) [111] (g) (111) (h) (011) (i) (030) () (121) (k) (113) (1) (071)arrow_forward
- In the following unit cells, which one has a vevtor pass through the cubic in direction [-2 -2 -1] ? 包,向 廊“廊, (a) (b) [V2 aarrow_forwardDetermine the angles a, ß, and y that are listed in the cubic unit cell provided. Enter the angles in degrees. Note: You should be able to use basic facts about cube geometry and crystallographic convention to solve this, rather than elaborate direction cosine equations. [00 1] (0 0 1) [1O 1] [11 1) (1 1 1) a = i B = iarrow_forwardwithin a orthorhombic unit cell, sketch the following: [213] and [301]arrow_forward
- 1 【Recommend】 Directly edit text and image in PDF Edit Question 1: University of Botswana Department of Mechanical Engineering Faculty of Engineering and Technology 2024/2025 session Semester I CCB 231/ CCB 211: Materials Science for Engineers Assignment I Due Date: 04/09/2024 Due Time: 14:00 Convert the following Miller indices to directions and sketch the directions within a cubic unit cell: (a) [301] (b) [301] (c) [301] (d) [122] (e) [122] [122] (g) [111] (h) [111] (i) [111] (j) [111] Question 2: The change of SO3 concentration with time is shown in Table 1. [C] t(s) 0 1.000000 100 0.778801 200 0.606531 300 0.472367 400 0.367879 500 0.286505 600 0.223130 800 0.135335 1000 0.082085 1200 0.049787 The process is governed by the rate law given by the equation: = In (SO3)-kt+In (SO3)-0 Where k is the rate constant and t is the time, and In is natural logarithm. By calculating the corresponding values of In[SO3] and plotting the graph of In [SO3] against time, t, determine the rate…arrow_forwarda: Sketch within a cubic unit cell the following crystallographic directions: [121], [201], [2 1 3), [111] [121] hi Dats JD fallarrow_forwardProblem #4 a) Find the indices for the directions indicated by the two vectors in the shown sketch below 0.5 mm 0.4 mm 0.7 mm Direction 1 Direction 2 ty b) Sketch the following directions within a cubic unit cell: (a) [123], (b) [211], (c) [102], (d) [133]. c) Sketch the following planes within a cubic unit cell: (a) (011), (b) (102), (c) (112), and (d) (131)arrow_forward
- I need the answer for (a) (b) and (c) Thanks..arrow_forwardThe attached photo shows the Young’s modulus of a cubic single crystal as a function of orientation. where a1 a2 and a3 are the direction cosines between the direction hkl and [100], [010], and [001], respectively. For a certain crystal, E111 = 500 GPa and E100 = 90 GPa. Calculate Young’s modulus for this single crystal in the <110> directionarrow_forwardNeed correctly with proper explanation. Will surely like it. Thank youarrow_forward
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