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Illusion [34]
2 years ago
12

1. Làm thế nào để đảm bảo tính khả thi của văn bản hành chính ?

Engineering
1 answer:
Dennis_Churaev [7]2 years ago
7 0

Answer:

Explanation:

......................

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What tool should be used to loosen or tighten brake or fuel lines?
emmainna [20.7K]

The tool you would use are brake line wrenches.

4 0
3 years ago
A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The oth
Lilit [14]

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

5 0
2 years ago
Why can't I fly. I dont know why?
algol13
In order to fly, you must have a device/mechanism that will release hot air, causing it to fly. A jet pack will do the job.
5 0
3 years ago
A hypothetical metal has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.413 nm, 0.665 nm, and 0.87
Inessa [10]

Answer:

atomic radius  R = 0.157 nm

metal atomic weight = 72.27 g/mol

Explanation:

given data

parameters a =  0.413 nm

parameters b = 0.665 nm

parameters c =  0.876 nm

atomic packing factor = 0.536

density = 3.99 g/cm³

to find out

atomic radius and  atomic weight

solution

we apply  here atomic packing factor (x) that is

atomic packing factor (x) = \frac{volume(sphere)}{volume(unit\ cell)}  ..................1

put here value we get

atomic packing factor = \frac{8*(4/3)*\pi R^3}{3*a*b*c}

R = (\frac{3(x)(abc)}{32\pi })^{1/3}

R =  (\frac{3(0.536)(0.413*0.665*0.876)}{32\pi })^{1/3}

atomic radius  R = 0.157 nm

and

now we get here metal atomic weight that is

metal atomic weight = \frac{\rho (abc)(N_A)}{no\ of\ atom}   ....................2

metal atomic weight = \frac{3.99 (0.413*0.665*0.876)(6.023*10^{23})}{8}  

metal atomic weight = 72.27 g/mol

6 0
3 years ago
The specifications of an electronic device are 24 /- 0.4 Amps. When the device exceeds specifications the average quality cost i
Misha Larkins [42]

Answer:

The average loss for this device is $10.90

Explanation:

Given data

Specification = 24 +/- 0.4 Amps

average quality cost = $32.00

average value y = 23.9

standard deviation s = 0.211 Amps

32 = k(24.4 - 24)²

32 = k(0.4)² = k(0.16)

k = 32/0.16 = 200

To evaluate average loss, use

L = k{s² + (y - T)²}

T = 24A

L = 200{0.211² + (23.9 - 24)²}

L = $10.90

The average loss for this device is $10.90

4 0
2 years ago
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