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Katyanochek1 [597]
3 years ago
5

What makes building an airplane while flying so difficult?

Engineering
1 answer:
Naddik [55]3 years ago
7 0

Answer: Because if something goes wrong while you are flying it it will crash

Explanation:

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A rotor in a compressor stage has a mean blade radius of 0.285 m and an angular rotor velocity of 8500 RPMs. The static temperat
pantera1 [17]

Answer:

0,285 is the answer

Explanation:

6 0
3 years ago
What should -7/56 BE DIVIDE TO GET -1/8​
aivan3 [116]

Answer:

7/7

Explanation:

7/56 ÷1

7/56÷ 7/7

7/56 ×7/7

1/8

5 0
3 years ago
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When we utilize a visualization on paper/screen, that visualization is limited to exploring: Group of answer choices Relationshi
Mila [183]

Answer:

As many variables as we can coherently communicate in 2 dimensions

Explanation:

Visualization is a descriptive analytical technique that enables people to see trends and dependencies of data with the aid of graphical information tools. Some of the examples of visualization techniques are pie charts, graphs, bar charts, maps, scatter plots, correlation matrices etc.

When we utilize a visualization on paper/screen, that visualization is limited to exploring as many variables as we can coherently communicate in 2-dimensions (2D).

6 0
3 years ago
Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and
Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in  

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒  ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that   PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

⇒  P = 4745.11 lb = 4.745 kips

The pic shown can help to understand the question.

5 0
3 years ago
A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per
mr_godi [17]

Answer:

-50.005 KJ

Explanation:

Mass flow rate = 0.147 KJ per kg

mass= 10 kg

Δh= 50 m

Δv= 15 m/s

W= 10×0.147= 1.47 KJ

Δu= -5 kJ/kg

ΔKE + ΔPE+ ΔU= Q-W

0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W

Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu

= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50

= 1.47 +3.375-4.8450-50

Q=-50.005 KJ

7 0
3 years ago
Read 2 more answers
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