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Mila [183]
3 years ago
10

This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1

.70), give formal proofs that the following languages are not regular: (a) L = {www | W € {0,1}* }. (b) L = {1"01" m, n >0}.
Engineering
1 answer:
Ulleksa [173]3 years ago
6 0

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>

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Answer:

e.Fire resistance,Inexpensive,Non-toxic.

Explanation:

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Answer:

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Explanation:

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7500 = ( F₁ - F₂ ) x 10

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750 = 2 F₂ - F₂      ( ∵F₁ = 2 F₂ )

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Now F₁ = 2 F₂

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Therefore we know,

F_{max} = 3 x F_{e}

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7 0
3 years ago
(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor
serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm =  0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

<u>A) would the 1040 steel be a possible candidate for this application</u>

<em>Yield strength of 1040 steel < stress  ( in order to be a possible candidate )</em>

stress = p / A0 = ( 18000 ) / ( \frac{\pi }{4} ) * 0.0075^2

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stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )  

Therefore 1040 steel is not a possible candidate for this application

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6 0
3 years ago
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Zinaida [17]

The load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

<h3>What is meant by torque?</h3>

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.

Let the beam is of length L

Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero

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also, we know that stress at both ends are same

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Now from two equations we have

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solving the above equation we have

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so the load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

The complete question is:

47. the beam is supported by two rods ab and cd that have cross-sectional areas of $$12mm^2 and $$8mm^2, respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.

To learn more about torque refer to:

brainly.com/question/20691242

#SPJ4

7 0
2 years ago
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