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dezoksy [38]
3 years ago
5

What is anthropology? Discuss the type of anthropology?

Engineering
2 answers:
Lisa [10]3 years ago
3 0

Anthropology is the study of humanity through the application of biology, cultural studies, archeology, linguistics and other social sciences.

Hope the attachment helps

disa [49]3 years ago
3 0

Answer:

action anthropology.

Explanation:

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I will definitely rate 5 stars/brainliest!!! HELP PLEASE!!! State University must purchase 1,100 computers from three vendors. V
romanna [79]
Why 1+12+ Y3 < 1100
Says the state of university Need to purchase 1100 computers in total, we have the following answer on the way top
3 0
3 years ago
A heat engine that rejects waste heat to a sink at 520 R has a thermal efficiency of 35 percent and a second- law efficiency of
xeze [42]

Answer:

The source temperature is 1248 R.

Explanation:

Second law efficiency of the engine is the ratio of actual efficiency to the maximum possible efficiency that is reversible efficiency.

Given:  

Temperature of the heat sink is 520 R.

Second law efficiency is 60%.

Actual thermal efficiency is 35%.

Calculation:  

Step1

Reversible efficiency is calculated as follows:

\eta_{II}=\frac{\eta_{a}}{\eta_{rev}}

0.6=\frac{0.35}{\eta_{rev}}

\eta_{rev}=0.5834

Step2

Source temperature is calculated as follows:

\eta_{rev}=1-\frac{T_{L}}{T}

\eta_{rev}=1-\frac{520}{T}

0.5834=1-\frac{520}{T}

T = 1248 R.

The heat engine is shown below:

Thus, the source temperature is 1248 R.

6 0
3 years ago
Air enters the 1 m² inlet of an aircraft engine at 100 kPa and 20° C with a velocity of 180 m/s. Determine: a) The volumetric fl
Shkiper50 [21]

Answer:

a) 180 m³/s

b) 213.4 kg/s

Explanation:

A_1 = 1 m²

P_1 = 100 kPa

V_1 = 180 m/s

Flow rate

Q=A_1V_1\\\Rightarrow Q=1\times 180\\\Rightarrow Q=180\ m^3/s

Volumetric flow rate = 180 m³/s

Mass flow rate

\dot{m}=\rho Q\\\Rightarrow \dot m=\frac{P_1}{RT} Q\\\Rightarrow \dot m=\frac{100000}{287\times 293.15}\times 180\\\Rightarrow \dotm=213.94\ kg/s

Mass flow rate = 213.4 kg/s

3 0
3 years ago
Suppose values is a sorted array of integers. Give pseudocode that describes how a new value can be inserted so that the resulti
Novosadov [1.4K]

Answer:

insert (array[] , value , currentsize , maxsize )

{

   if maxsize <=currentsize

  {

      return -1

  }

  index = currentsize-1

  while (i>=0 && array[index] > value)

  {

      array[index+1]=array[index]

      i=i-1

  }

 

  array[i+1]=value

  return 0

}

Explanation:

1: Check if array is already full, if it's full then no component may be inserted.

2: if array isn't full:

  • Check parts of the array ranging from last position of range towards initial range and determine position of that initial range that is smaller than the worth to be inserted.  
  • Right shift every component of the array once ranging from last position up to the position larger than the position at that smaller range was known.
  • assign new worth to the position that is next to the known position of initial smaller component.
7 0
3 years ago
The time factor for a doubly drained clay layer
Margarita [4]

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

T_v=\frac{\pi }{4}(\frac{U}{100})^2

Solving for 'U' we get

\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i)\frac{z}{H}=0.25=U=0.71 = 71% consolidation

ii)\frac{z}{H}=0.5=U=0.45 = 45% consolidation

iii)\frac{z}{H}=0.75U=0.3 = 30% consolidation

Part b)

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm

5 0
3 years ago
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