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Alekssandra [29.7K]
2 years ago
11

Define ways in which you would go about networking to explore opportunities in your career field and obtain more information for

yourself. If applicable, explain ways in which you have already done that and how you will expand going forward. Choose one person in your career field that you would like to have a conversation with. What do you think you can learn from this person?
Engineering
1 answer:
kkurt [141]2 years ago
6 0

Answer:

mmmmmm

Explanation:

ffmfmfmmfmfmmfmfmfmfmfmffmfmfmfmfmfmfmfmfmfmfmfmfffmmfmffmmmfmfmfmfmfmfmfmfffmfmfmfmfmfmmfmfmmfmfmfmfmfmfffmfmfffmfmfmfmffmmfmfmffmfmfmfmfffmfmfmfmfmmmmfmfmfmfmfmfmfmmmfmfmfmfmfmfmfmfmmfmfmfmfmfmfmfmfmfmfmffmfmfmfmfmfmmfmfmmmmmfmfmfmffmfmfmfmffmffmfm

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Which type of engineer is needed in the following scenario?
gregori [183]

Answer:

A: Agricultural Engineer

Explanation:

I had this same question for a test and got it right with a being the answer :)

8 0
2 years ago
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Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450 K, 350 kPa, and velocity of 3
viktelen [127]

Answer:

Given that

Mass flow rate ,m=2.3 kg/s

T₁=450 K

P₁=350 KPa

C₁=3 m/s

T₂=300 K

C₂=460 m/s

Cp=1.011 KJ/kg.k

For ideal gas

P V = m R T

P = ρ RT

\rho_1=\dfrac{P_1}{RT_1}

\rho_1=\dfrac{350}{0.287\times 450}

ρ₁=2.71 kg/m³

mass flow rate

m= ρ₁A₁C₁

2.3 = 2.71 x A₁ x 3

A₁=0.28 m²

Now from first law for open system

h_1+\dfrac{C_1^2}{200}+Q=h_2+\dfrac{C_2^2}{2000}

For ideal gas

Δh = CpΔT

by putting the values

1.011\times 450+\dfrac{3^2}{200}+Q=1.011\times 300+\dfrac{460^2}{2000}

Q=1.011\times 300+\dfrac{460^2}{2000}-\dfrac{3^2}{200}-1.011\times 450

Q= - 45.49 KJ/kg

Q =- m x 45.49 KW

Q= - 104.67 KW

Negative sign indicates that heat transfer from air to surrounding

4 0
2 years ago
Does anybody want 20 points? They're free get 'em while ya can...
DedPeter [7]

Answer:

hi

Explanation:

4 0
2 years ago
Read 2 more answers
If a system of pulleys results in a force of 25% of the load, how far will the rope need to move to pull the load a distance of
GaryK [48]

Answer:

  40 ft

Explanation:

Assuming no loss of energy in the system of pulleys, the work done is the same whether you move the load directly or through the pulleys.

  W = Fd . . . . . . . . work is the product of force and distance

  F(10 ft) = (0.25F)(d) . . . . . where d is the distance we want to find

  d = 10F/(0.25F) = 40

The rope will need to move 40 feet.

8 0
2 years ago
Water flows with an average speed of 6.5 ft/s in a rectangular channel having a width of 5 ft The depth of the water is 2 ft.
lisov135 [29]

Answer:

specific energy  = 2.65 ft

y2 = 1.48 ft  

Explanation:

given data

average speed v = 6.5 ft/s

width = 5 ft

depth of the water y = 2 ft

solution

we get here specific energy that is express as

specific energy = y + \frac{v^2}{2g}     ...............1

put here value and we get

specific energy = 2 + \frac{6.5^2}{2\times 9.8\times 3.281}  

specific energy  = 2.65 ft

and

alternate depth is

y2 = \frac{y1}{2} \times (-1+\sqrt{1+8Fr^2})  

and

here Fr² = \frac{v1}{\sqrt{gy}}  = \frac{6.5}{\sqrt{32.8\times 2}}  

Fr² = 0.8025

put here value and we get

y2 = \frac{2}{2} \times (-1+\sqrt{1+8\times 0.8025^2})

y2 = 1.48 ft  

7 0
3 years ago
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