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3 years ago

What is anthropology? Discuss the type of anthropology?

2 answers:

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