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rusak2 [61]
3 years ago
9

Please help me: Use the Node analysis to find the power of all resistors

Engineering
1 answer:
Viktor [21]3 years ago
3 0
‘Politics and planning are increasingly gaining prominence in contemporary urban and regional planning debates’. Using relevant examples, discuss this assertion reflecting on the critical success factors for the successful implementation of the land reform program in South Africa.
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The steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. If the valve A is closed and the water pre
xenn [34]

Find the below attachment

8 0
3 years ago
A heat engine operates between a source at 477°C and a sink at 27°C. If heat is supplied to the heat engine at a steady rate of
lara [203]

Answer:

T_C = 27+273.15 = 300.15 K

T_H = 477+273.15 = 750.15 K

And replacing in the Carnot efficiency we got:

e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%

W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}

Explanation:

For this case we can use the fact that the maximum thermal efficiency for a heat engine between two temperatures are given by the Carnot efficiency:

e = 1 -frac{T_C}{T_H}

We have on this case after convert the temperatures in kelvin this:

T_C = 27+273.15 = 300.15 K

T_H = 477+273.15 = 750.15 K

And replacing in the Carnot efficiency we got:

e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%

And the maximum power output on this case would be defined as:

W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}

Where Q_H represent the heat associated to the deposit with higher temperature.

4 0
3 years ago
The cylinder C is being lifted using the cable and pulley system shown.
Sonja [21]

Answer:

sry but it's kinda hard

4 0
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
A TV USE 75 WATTS WHILE IN USED ASSMING THAT ITIS USED 4 HOURS EVERY DAY HOW MUCH ENERGY IN 4 IN KWH WOULD THE TV CONSUME ANNUAL
prohojiy [21]

Answer:

i don't think i understand the question

Explanation:

7 0
2 years ago
Read 2 more answers
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