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ira [324]
4 years ago
12

At 120 KTAS, what is a good angle of bank to use to generate a level standard rate turn for instrument flight conditions

Engineering
1 answer:
wlad13 [49]4 years ago
3 0

Answer:

The correct answer will be "18 degrees AOB".

Explanation:

The given value is:

KTAS = 120

For determining bank angles through standard rate switches, as we understand

⇒  10 \ percent \ of KTAS +\frac{1}{2} \ of \ the \ number

On substituting the given value of KTAS, we get

⇒  10 \ percent \ of 120+\frac{1}{2} \ of \ the \ number

⇒  12+6

⇒  18 \ degree \ angle \ of \ bank

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QPSK: 7.5 MHz

64-QAM:2.5 MHz

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Explanation:

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A gas refrigeration system using air as the working fluid has a pressure ratio of 5. Air enters the compressor at 0°C. The high-
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3 years ago
Courtney spent $87 at a sporting good store. She spent $29 on a tennis racket and the rest on new shoes How much money did Court
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3 years ago
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A manager has a list of items that have been sorted according to an item ID. Some of them are duplicates. She wants to add a cod
ruslelena [56]

Answer:

The solution code is written in Python:

  1. items = [{"id": 37697, "code": ""},{"id": 37698, "code": ""},{"id": 37699, "code": ""},{"id": 37699, "code": ""}, {"id": 37699, "code": ""},
  2. {"id": 37699, "code": ""},{"id": 37699, "code": ""},{"id": 37699, "code": ""},{"id": 37700, "code": ""} ]
  3. items[0]["code"] = 1
  4. for i in range(1, len(items)):
  5.    if(items[i]["id"] == items[i-1]["id"]):
  6.        items[i]["code"] = items[i-1]["code"] + 1
  7.    else:
  8.        items[i]["code"] = 1
  9. print(items)

Explanation:

Firstly, let's create a list of dictionary objects. Each object holds an id and a code (Line 1-2). Please note all the code is initialized with zero at the first beginning.

Next, we can assign 1 to the <em>code</em> property of items[0] (Line 4).

Next, we traverse through the items list started from the second element (Line 6). We set an if condition to check if the current item's id is equal to the previous item (Line 7). If so, we assign the previous item's code + 1 to the current item's code (Line 8). If not, we assign 1 to the current item's code (Line 10).

At last, we print out the item (Line 12) and we shall get

[{'id': 37697, 'code': 1}, {'id': 37698, 'code': 1}, {'id': 37699, 'code': 1}, {'id': 37699, 'code': 2}, {'id': 37699, 'code': 3}, {'id': 37699, 'code': 4}, {'id': 37699, 'code': 5}, {'id': 37699, 'code': 6}, {'id': 37700, 'code': 1}]

 

6 0
4 years ago
Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical
Olin [163]

Answer:

The theoretical density for Niobium is 1.87 g/cm^3.

Explanation:

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density  of the unit cell

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

We have :

Z = 2 (BCC)

M = 92.91 g/mol ( Niobium)

Atomic radius for niobium = r = 0.143 nm

Edge length of the unit cell = a

r = 0.866 a (BCC unit cell)

a=\frac{0.143 nm}{0.866}=0.165 nm=0.165 \times 10^{-7} cm

1 nm = 10^{-7} cm

On substituting all the given values , we will get the value of 'a'.

\rho=\frac{2\times 92.91}{6.022\times 10^{23} mol^{-1}\times (0.165 \times 10^{-7} cm)^{3}}

\rho =1.87 g/cm^3

The theoretical density for Niobium is 1.87 g/cm^3.

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3 years ago
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