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swat32
3 years ago
12

The best penetration is achieved with DCEN current. True or false

Engineering
1 answer:
GenaCL600 [577]3 years ago
6 0

Answer:

True

Explanation:

The best penetration is achieved with dcen current. Mild steel is expensive and requires the most amount of cleaning. Copper-coated steel welding rods are not used for gas tungsten arc welding because they will continue the weld or electrode.

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The flowchart below shows the design steps required to build a working model.
AlekseyPX

Question:

1) test the model and analyze the results of the test

2) building the model and observing it

3) observing the model and reporting results

w

4) designing the model and drawing conclusions​

Answer:

The correct option is;

1) Test the model and analyze the results of the test

Explanation:

Based on the flowchart, a model improvement process involves the implementation of a process or model improvement cycle such as the Plan Do Study Act, PDSA cycle, however feedback to the process will be be gotten from testing the model and analyzing the results of the tests. When grey areas or aspects of the model are found that cause the system to malfunction are determined, steps should then be taken to improve the performance of the model.

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3 years ago
List the thermal conductivities of five
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Answer:

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Explanation:

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Using the tables for water, determine the specified property data at the indicated states. In each case, locate the state on ske
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3 years ago
Given two alphabet strings str1 and str2. You can change the characters in str1 to any alphabet characters in order to transform
Keith_Richards [23]

Answer:

Explanation:

5

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3 years ago
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
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