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Fudgin [204]
2 years ago
12

Câu 1: Tìm từ trong 2 câu thơ sau liên quan đến các từ: hình chiếu, tia chiếu, mặt phẳng chiếu.

Engineering
1 answer:
xeze [42]2 years ago
4 0

Explanation:

are bhai brainly aap english me questions kar ne ko hai

ye kon si bhasha hai ??????

You might be interested in
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|

5 0
3 years ago
Who is/are the founder/founders of transistor? ​
den301095 [7]

Answer:

William Shockley, Walter Houser Brattain and John Bardeen.

Explanation:

It was built in 1947 and they won the novel peace prize in 1956

7 0
3 years ago
Read 2 more answers
What do Engineers aim to do
Basile [38]

Answer:

Engineers design, evaluate, develop, test, modify, install, inspect and maintain a wide variety of products and systems.

Explanation:

6 0
2 years ago
Read 2 more answers
A missile flying at high speed has a stagnation pressure and temperature of 5 atm and 598.59 °R respectively. What is the densit
alexdok [17]

Answer:

5.31\frac{kg}{m^3}

Explanation:

Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to \frac{m}{M}, where m is the mass and M the molar mass of the gas, and the density is \frac{m}{V}.

For air M=28.66*10^{-3}\frac{kg}{mol} and \frac{5}{9}R=K

So, 598.59 R*\frac{5}{9}=332.55K

pV=nRT\\pV=\frac{m}{M}RT\\\frac{m}{V}=\frac{pM}{RT}\\\rho=\frac{pM}{RT}\\\rho=\frac{(5atm)28.66*10^{-3}\frac{kg}{mol}}{(8.20*10^{-5}\frac{m^3*atm}{K*mol})332.55K}=5.31\frac{kg}{m^3}

7 0
3 years ago
A walrus loses heat by conduction through its blubber at the rate of 220 W when immersed in −1.00°C water. Its internal core tem
Llana [10]

Answer:

The average thickness of the blubber is<u> 0.077 m</u>

Explanation:

Here, we want to calculate the average thickness of the Walrus blubber.

We employ a mathematical formula to calculate this;

The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L

Where dQ is the change in amount of heat transferred

dT is the temperature gradient(change in temperature) i.e T2-T1

dQ/dT = 220 W

K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)

A is the surface area which is 2.23 m^2

T2 = 37.0 °C

T1 = -1.0 °C

L is ?

We can rewrite the equation in terms of L as follows;

L × dQ/dT = KA(T2-T1)

L = KA(T2-T1) ÷ dQ/dT

Imputing the values listed above;

L = (0.2 * 2.23)(37-(-1))/220

L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m

7 0
3 years ago
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