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spin [16.1K]
3 years ago
14

Which is not an affect of alcohol

Engineering
1 answer:
Gre4nikov [31]3 years ago
6 0
Pooping problems is not an affect
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
A 400-m^3 storage tank is being constructed to hold LNG, liquefied natural gas, which may be assumed to be essentially pure meth
GuDViN [60]

Answer:

mass of LNG: 129501.3388 kg

quality: 0.005048662

Explanation:

Volume occupied by liquid:

400 m^3*0.9 = 360 m^3

Volume occupied by vapor

400 m^3*0.1 = 40 m^3  

A figured with thermodynamic properties of saturated methane is attached. Notice that a liquid-gas mixture is present

For liquid phase specific volume (vf) at 150 K is 0.002794 m^3/kg and for vapor phase specific volume (vg) is 0.06118 m^3/kg

From specific volume definition:

vf = liquid volume/liquid mass

liquid mass = liquid volume/vf

liquid mass = 360 m^3/0.002794 m^3/kg

liquid mass = 128847.5304 kg

vg = vapor volume/vapor mass

vapor mass = liquid volume/vg

vapor mass = 40 m^3/0.06118 m^3/kg

vapor mass = 653.8084341 kg

total mass = 128847.5304 kg + 653.8084341 kg = 129501.3388 kg

Quality is defined as the ratio between vapor mass and total mass

quality =  653.8084341 kg/129501.3388 kg = 0.005048662

4 0
3 years ago
How do you identify all sensors, functions, and where we can use them?
Alex17521 [72]

Sensor/Detectors/Transducers are electrical, opto-electrical, or electronic devices composed of specialty electronics or otherwise sensitive materials, for determining if there is a presence of a particular entity or function. Many vehicles including cars, trains, buses etc. all use sensors to monitor oil temperature and pressure, throttle and steering systems and so many more aspects.

7 0
3 years ago
A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m
natima [27]

Answer:

a. 0.4544 N

b. 5.112 \times 10^{-5 M}

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O

NaOH\ Mass = Normality \times equivalent\ weight \times\ volume

= 0.3200 \times 40 g \times 21.30 mL \times  1L/1000mL

= 0.27264 g

NaOH\ mass = \frac{mass}{molecular\ weight}

= \frac{0.27264\ g}{40g/mol}

= 0.006816 mol

Now

Moles of H_2SO_4 needed  is

= \frac{0.006816}{2}

= 0.003408 mol

Mass\ of\ H_2SO_4 = moles \times molecular\ weight

= 0.003408\ mol \times 98g/mol

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

= \frac{acid\ mass}{equivalent\ weight \times volume}

= \frac{0.333984 g}{49 \times 0.015}

= 0.4544 N

b. And, the acid solution molarity is

= \frac{moles}{Volume}

= \frac{0.003408 mol}{15\ mL \times  1L/1000\ mL}

= 0.00005112

=5.112 \times 10^{-5 M}

We simply applied the above formulas

4 0
3 years ago
Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation t
densk [106]

Answer:

hello your question is incomplete attached below is the complete question

answer : attached below

Explanation:

let ; x(t)  be a real value signal for x ( jw ) = 0 , |w| > 200\pi

g(t) = x ( t ) sin ( 2000 \pi t )

x_{1} (t) = \frac{1}{2}  x(t)  sin ( 4000\pi t )

next we apply Fourier transform

attached below is the remaining part of the solution

6 0
3 years ago
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