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Vitek1552 [10]
3 years ago
14

Can some please help me with my questions. The Questions are in these two pictures below. I don't know how to do this question.

Engineering
1 answer:
ololo11 [35]3 years ago
5 0

Answer: ok if you need help go to help me with a question.com

Explanation:

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Find the general solution of the equation<br>a) Tan A = 1/√3​
castortr0y [4]

Given :

tan \ A = \dfrac{1}{\sqrt{3}}

To Find :

The general solution of the given equation.

Solution :

We know, tan\ 30^o = \dfrac{1}{\sqrt{3}}

Comparing it with given equation, we get :

A = \dfrac{\pi}{6}

General solution is given by :

A = n\pi  + \dfrac{\pi}{6} where n ∈ Z ( integers )

Hence, this is the required solution.

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3 years ago
Whose responsibility is it to provide direction on correct ladder usage?<br> select the best option.
viva [34]

Answer:

it is the employers responsibility to provide direction on Correct ladder usage.

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2 years ago
Time complexity of merge sort
vovangra [49]

Answer:

The correct answer is "O (n\times Log n)". A further explanation is given below.

Explanation:

  • Throughout all the three instances (worst, average as well as best), the time complexity including its Merge sort seems to be O (n\times Log n) as the merge form often splits the array into two halves together tends to linear time to combine multiple halves.
  • As an unsorted array, it needs an equivalent amount of unnecessary capacity. Therefore, large unsorted arrays are not appropriate for having to search.
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3 years ago
Please help me in this assignment.
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3 years ago
A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 240 kJ of work on the
Sergeu [11.5K]

To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.

By definition the entropy change would be defined as

\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})

Using the Boyle equation we have

\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})

Where,

C_p = Specific heat at constant pressure

T_1= Initial temperature of gas

T_2= Final temprature of gas

R = Universal gas constant

v_1= Initial specific Volume of gas

v_2= Final specific volume of gas

According to the statement, it is an isothermal process and the tank is therefore rigid

T_1 = T_2, v_2=v_1

The equation would turn out as

\Delta S = C_p ln1-ln1

<em>Therefore the entropy change of the ideal gas is 0</em>

Into the surroundings we have that

\Delta S = \frac{Q}{T}

Where,

Q = Heat Exchange

T = Temperature in the surrounding

Replacing with our values we have that

\Delta S = \frac{230kJ}{(30+273)K}

\Delta S = 0.76 kJ/K

<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>

8 0
3 years ago
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