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3 years ago

Match the description with the term. I need help

Engineering

1 answer:

mel-nik [20]3 years ago

6 0

Answer:

cultivation - preparing and planting crops

domestication - capturing, taming, and breeding animals

hunting and gathering - obtaining food from the wild

Explanation:

moo

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What is the differences between stack and queue?

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A stack is an ordered list of elements where all insertions and deletions are made at the same end, whereas a queue is exactly the opposite of a stack.

Explanation:

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3 years ago

What does this map show about the Viking civilization?

BlackZzzverrR [31]

Answer:

the answer is c

Explanation:

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3 years ago

Read 2 more answers

A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an

Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here thermodynamic equation that

energy equation that is

$h1 + \frac{v1}{2} + q = h2 + \frac{v2}{2} + w$

put here value with

turbine is insulated so q = 0

so here

$3015.4 *1000 + \frac{10^2}{2} = 2431.7 * 1000 + \frac{90^2}{2} + w$

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

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3 years ago

Which of the following is most useful for doing research?

Ghella [55]

Answer:

Web Browser

Explanation:

Because you dont use a messaging app or presentation software to look up stuff its common knowledge

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2 years ago

A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

2 years ago