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bearhunter [10]
3 years ago
15

Anything you want to do in Hootsuite can be found in the ________, with the main workspace in the _________?

Engineering
1 answer:
Thepotemich [5.8K]3 years ago
5 0

Answer:

Anything you want to do in Hootsuite can be found in the ___ Sidebar_____, with the main workspace in the ___center______

Sidebar; center

Explanation:

The main workspace of the Hootsuite is located in the center.  The sidebar is where the core access to the Hootsuite functionality, like Streams, Inbox, Planner, Analytics, Publisher, and the App Directory, is obtained.  Hootsuite is a media management platform for curating content, scheduling posts, managing team members, and measuring performances.

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Miller Indices:
svetlana [45]

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.

<u>In this case, for [1 2 1]:</u>

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

<u>for </u>[1 2 \overline{4}]<u>:</u>

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.

<u>In this case, for [1 2 1]:</u>

<u />d_{121}= \displaystyle \frac{a}{\sqrt{1^2+2^2+1^2}}=\frac{a}{\sqrt{6}}=0.41a<u />

<u>for </u>[1 2 \overline{4}]<u>:</u>

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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3 years ago
What is a voltage divider circuit and how do you calculate the voltage across one element in a series
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2 years ago
Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss
klemol [59]

Answer:

Heat gain of 142 kJ

Explanation:

We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.

ΔU = Q + W ⇒ Q = ΔU -W

Q = 79 - (-63) = 142 kJ

Therefore, the gas gained heat by an amount of 142 kJ.

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