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Sever21 [200]
3 years ago
15

In the long run, if the firm decides to keep output at its initial level, what will it likely do? Stay on SRATC3 but decrease to

the point touching LRATC Shift to operate on SRATC4 Shut down Shift to operate on SRATC2 True or False: The minimum efficient scale is achieved at the minimum point on each average total cost curve. False True
Engineering
1 answer:
romanna [79]3 years ago
8 0

Answer:

Stay on SRATC3 but decrease to the point touching LRATC.

False

Explanation:

The long run average total cost is the total cost for the firm to continue its operations. In the given scenario the firm decides to keep its level of out =put at initial level then it should stay in short run average total cost and then gradually moving towards long run average total cost.

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Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of shaft power. The free surface of t
ahrayia [7]

Answer:

note:

<u><em>solution is attached due to error in mathematical equation. please find the attachment</em></u>

8 0
2 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
10. Identify one material we Mine and what we make with that material
Katen [24]

Answer:

We Mine Limestone. Its used to make cement, toothpaste or paints, soil conditioner, and rip rap stone.

Explanation:

7 0
2 years ago
A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twi
Leviafan [203]

Answer:

F₁ = 1500 N

F₂ = 750 N

F_{e} = 500 N

Explanation:

Given :

Power transmission, P = 7.5 kW

                                      = 7.5 x 1000 W

                                      = 7500 W

Belt velocity, V = 10 m/s

F₁ = 2 F₂

Now we know from power transmission equation

P = ( F₁ - F₂ ) x V

7500 = ( F₁ - F₂ ) x 10

750 =  F₁ - F₂

750 = 2 F₂ - F₂      ( ∵F₁ = 2 F₂ )

∴F₂  = 750 N

Now F₁ = 2 F₂

        F₁ = 2 x F₂

        F₁ = 2 x 750

        F₁ = 1500 N   ,   this is the maximum force.

Therefore we know,

F_{max} = 3 x F_{e}

where F_{e} is centrifugal force

 F_{e} = F_{max} / 3

                          = 1500 / 3

                         = 500 N

8 0
3 years ago
To prevent drainage of the transmission fluid from the converter when the
-BARSIC- [3]
Answer
D I think
Explanation
4 0
2 years ago
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