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lbvjy [14]
3 years ago
7

Can you answer what is attached.

Engineering
1 answer:
kirill [66]3 years ago
3 0
29.4 bro I hope that helps
You might be interested in
This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1
Ulleksa [173]

Answer:

<em>L is not a regular language with formal proofs  </em>

Explanation:

<em>(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails  that L is a regular language. Then by the Pumping Lemma for Regular Languages, </em>

<em>there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, </em>

<em>s = xyz subject to the following conditions: </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L</em>

<em />

<em>(b) To determine that L is not a regular language, we mke use of proof by contradiction.  lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,</em>

<em>The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject  to the condtions as follows : </em>

<em>(a) |y| > 0 </em>

<em>(b) |xy| ≤ p, and </em>

<em>(c) ∀i > 0, xyi </em>

<em>z ∈ L. </em>

<em>Choose s = 0p10p </em>

<em>. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.</em>

<em>for some  k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus,  xy0 </em>

<em>z should be in L. xy0 </em>

<em>z = xz = 0(p−k)10p </em>

<em>It is shown that is is  not in L. This is a  contraption with the pumping lemma.  our assumption that L is regular is  incorrect, and L is not a regular language</em>

6 0
3 years ago
What are the characteristic features of stress corrosion cracks?
Oduvanchick [21]

Answer and Explanation:

The crack formation growth that takes place in an environment corrosive.

Stress corrosion cracks can be defined as the spontaneous failures of the metal alloy as a result of the combined action of corrosion and high tensile stress.

Some of the characteristic features of stress corrosion cracks are:

  • These occur at high temperatures.
  • Occurrence of failures in metals mechanically.
  • Occurrence of sudden and unexpected failures under tensile stress.
  • The rate of work hardening of the metal alloy is high.
  • Time
  • An environment that is specific for stress corrosion cracking.
5 0
3 years ago
Which two is right about febuary 14
igor_vitrenko [27]

Answer:A and B

Explanation:

3 0
3 years ago
Read 2 more answers
The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100),
skelet666 [1.2K]

Answer:

a)5.28 Å , b)3.73 Å , c)3.048 Å

Explanation:

the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.

Therefore, a particular unit cell consist only 1/8th part of an atom.

The lattice constant of a simple cubic primitive cell is 5.28 Å

We know formula of distance,

d = \frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}

a)(100)

a=5.28 Å

Distance = \frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}=5.28 Å

b)(110)

Distance = \frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}} = 3.73 Å

c)(111)

Distance= \frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}= 3.048 Å

6 0
3 years ago
Find the Rectangular form of the following phasors?
almond37 [142]

Answer:

The angles are missing in the question.

The angles are :

45,     30,    60,     90,    -34,     -56,      20,     -42,  -65,    -15

P=10, P=5,  P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10,   θ = 45°  rectangular coordinates

x = r cosθ  ,   y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

  = 7.07

y =P sinθ = 10 sin 45°

  = 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos  30° = 4.33

y = 5 sin  30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos  60° = 12.5

y = 25 sin  60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos  90° = 0

y = 54 sin  90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos  (-34°) = 53.88

y = 65 sin  (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos  (-56)° = 53.12

y = 95 sin  (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos  20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos  (-42)° = 5.94

y = 8 sin  (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos  (-65)° = 14.79

y = 35 sin  (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos  (-15)° = 144.88

y = 150 sin  (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

6 0
3 years ago
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