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VLD [36.1K]
3 years ago
14

If a sky diver decides to jump off a jet in Arkansas

Engineering
1 answer:
andrew-mc [135]3 years ago
6 0

Answer:

The answer to this question can be defined as follows:

Explanation:

The sky driver began his sky journey from Arkansas, drove across the Tennessee River then landed in North Carolina. He returned to both the north in the very same direction. He began with NC, traveled through Tennessee, eventually lands in Arkansas. But North Carolina has been in the third state on which skydiver was traveling over, and It's also more than 700 miles from Arkansas to the NC.

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Polymers can be natural or synthetic. a)-True b)- false?
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TRUE

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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
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