Answer:
A)The sketches for the required planes were drawn in the first attachment.
B)The sketches for the required directions were drawn in the second attachment.
To draw a plane in a simple 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)
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.
To draw a direction in a simple cubic lattice, you have to follow these instructions:
1- Identify the points a₀, b₀, and c₀ in the cubic cell.
2- Draw the direction as a vector-like (a₀ b₀ c₀).
Answer:
156.0 ksi
Explanation:
The formula of compressive strength is CS = F / A
where F is the force or load while A is the cross-sectional area
Given the inertia Ag = 35.1 in^4 , which is less than 40, the steel has a short column.
E = 29000 ksi
L = 12ft
r = 2.69 in
therefore;
CS = π^2 E / (kL^2/r)^2
= π^2 29000 / (0.8 x 12^2 / 2.69)^2
= 286292.756 / 1834.4089 = 156 ksi
Answer:
W = 1 mJ
V_new = 2000 V
W_new = 2 mJ
The extra energy came from the work done from moving the plates
Explanation:
We are given;
Capacitance; C = 1000pF = 10^(-9) F
Voltage; V = 1000V
Now,formula for stored energy in a parallel plate capacitor is given by;
W = (1/2)CV² = (1/2)(10^(-9))(1000²) = 0.5 mJ
However, in this case, it's W = 0.5 x 2 = 1 mJ since parallel-plate capacitor with air dielectric
When plates are moved and distance between plates is doubled(net charge is the same), thus we can calculate voltage from;
Q = CV
Since, C_new = C/2
Thus,
Q = (C/2)V_new
V_new = 2Q/C
Thus, V_new = 2V
Thus, V_new = 2 x 1000 = 2000 V
Now,
W_new = (1/2)C_new•(V_new)² = (1/2) (0.5C)•(2V)² = CV² = 2W = 2 x 1 = 2mJ
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