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Anna71 [15]
3 years ago
11

The themes around which social sciences texts are organized boost understanding by

Engineering
1 answer:
Nadya [2.5K]3 years ago
7 0

Answer:

  • <em>Facilitating an immensely focused setting where one does not get carried away </em>
  • <em>Creates an easier means to develop essays in relation to targeted themes </em>
  • <em>Developing "umbrellas" that one can create sub groups that a reader comes across </em>
  • <em>Instilling a directive path for vocabulary reading</em>
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The Clausius inequality expresses which of the following laws? i. Law of Conservation of Mass ii. Law of Conservation of Energy
DanielleElmas [232]

Answer:

(iv) second law of thermodynamics

Explanation:

The Clausius  inequality expresses the second law of thermodynamics it applies to the real engine cycle.It is defined as the cycle integral of change in entropy of a reversible system is zero. It is nothing but mathematical form of second law of thermodynamics . It also states that for irreversible process the cyclic integral of change in entropy is less than zero

3 0
3 years ago
Why must air tanks be drained​
Jobisdone [24]
Water can freeze in cold weather and cause brake failure.
7 0
3 years ago
Entropy change is evaluated using Eq. 6.2a based on an internally reversible process. Can the entropy change between two states
Vadim26 [7]

Answer:

YES

Explanation:

Entropy is an extensive property of the system entropy change that value of entropy change can be determined for any process between the states whether reversible or not. i have attached the formula to calculate entropy change which is independent of whether the system is reversible or not and can be determined for any process.

4 0
3 years ago
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

6 0
2 years ago
List all possible fracture mechanisms under which the unidirectional composites fail. Briefly explain and describe the related m
professor190 [17]

Answer:

Ususushehehehhuuiiïbbb

Explanation:

Yyshehshehshshsheyysysueueue

7 0
2 years ago
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