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rodikova [14]
3 years ago
7

Which of these is an example of a service job?

Engineering
1 answer:
Sindrei [870]3 years ago
7 0

Answer:

brainllest if right

A

Explanation:

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When I was taking my sat exam online and my phone battery died

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If an imbalance occurs, the _
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A. AFGI is the answer for this question.

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A normal shock wave takes place during the flow of air at a Mach number of 1.8. The static pressure and temperature of the air u
Darina [25.2K]

Answer:

The pressure upstream and downstream of a shock wave are related as

\frac{P_{1}}{P_{o}}=\frac{2\gamma M^{2}-(\gamma -1)}{\gamma +1}

where,

\gamma= Specific Heat ratio of air

M = Mach number upstream

We know that \gamma _{air}=1.4

Applying values we get

\frac{P_{1}}{100kPa}=\frac{2\times 1.4\times 1.8^{2}-(1.4 -1)}{1.4 +1}\\\\\frac{P_{1}}{100kPa}=3.61\\\\\therefore P_{1}=361.33kPa(Absloute)

Similarly the temperature downstream is obtained by the relation

\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}

Applying values we get

\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C

The Mach number downstream is obtained by the relation

M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616

3 0
4 years ago
Estimate horizontal and vertical tail sizes fo r a twin-turboprop regional transport with wing area of 40 m2, aspect ratio of 10
svetlana [45]

Answer: 0.2m sqr

Explanation:

A well behaved aircraft basically have a value of volume in horizontal and vertical area.

Volume in horizontal area (Vh) = 0.6

Volume in vertical area (Vv) = 0.05

Having known this, consider the relationship to find the vertical and horizontal tail sizes.

Vertical tail area (Sv)

Horizontal tail area (Sh)

Vh= (Sh × I) / S

Where,

I = moment

S= wing area

Sh= Horizontal tail area

Vh= Volume in horizontal area

0.6= Sh × 10/40

24= 10Sh

Sh= 24/10

Sh= 2.4 msqr

Horizontal tail area= 2.4m sqr

From the information above, we can calculate the vertical tail area.

Vertical tail area is calculated thus below:

Vv= (Sv× I) / S

Where

Vv= Volume in vertical area

Sv= Vertical tail area

I= Moment

S= Wing area

Therefore

Sv= (Vv × S) /I

Sv= (0.05×40)/10

Sv= 0.2msqr

In conclusion, the vertical tail size is 0.2msqr

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3 years ago
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Which option identifies the type of device the engineer will develop in the following scenario?
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It would be actuator
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