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Feliz [49]
3 years ago
6

Although it shouldn’t have happened, on a dive i fail to watch my spg and run out of air. if my buddy is close by, my best optio

n is to another option is to if i’m in shallow water and the surface is closer than my buddy.a) ascend using my buddy's alternate air source / make a controlled emergency swimming ascent
b) ascend using my buddy's alternate air source / make an buoyant emergency ascent
c) make a normal ascent / ascend using my buddy's alternate air source
d) make a controlled emergency swimming ascent / make a normal ascent
Physics
2 answers:
sukhopar [10]3 years ago
8 0

Complete Question:

Although it shouldn't have happened, on a dive I fail to watch my SPG and run out of air. If my buddy is close by, my best option is to __________. Another option is to _____________, if I'm in shallow water and the surface is closer than my buddy.

NB: The options still remain as written in the question

Answer:

a) ascend using my buddy's alternate air source / make a controlled emergency swimming ascent

Explanation:

Since I have exhausted my own air source, my buddy's alternate air source will be the only saver if he is close by. Controlled emergency swimming accent is used if the swimmer's buddy is far but the swimmer is close to the surface. Buoyant emergency ascent can only be used if the swimmer is deep away from the surface and needs to make it to the surface.

leva [86]3 years ago
5 0

Answer:

B ) Ascend using my buddy alternative air source / make an emergency Ascent

Explanation:

From the description it can be seen his buddy is close by of which he can easily use the alternative air source. Also we can see that he is closer to the water surface than his buddy, of which controlled emergency swimming ascent is highly favourable in this condition.

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This question is incomplete, the complete question is;

A rocket engine has a chamber pressure 4 MPa and a chamber temperature of 2000 K. Assuming isentropic expansion through the nozzle, and an exit Mach number of 3.2, what are the stagnation pressure and temperature in the exit plane of the nozzle?  Assume the specific heat ratio is 1.2.

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- the stagnation temperature Tt is 4048 K

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Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1) = ( Tt/T )^(γ/γ -1)

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we substitute

Pt/P = (1 + (γ-1 / 2) M²)^(γ/γ -1)

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Pt/P = ( Tt/T )^(γ/γ -1)

we substitute

274.993 × 10⁶ Pa / 4 × 10⁶ Pa =  ( Tt / 2000 )^(1.2/1.2 -1)

68.7484 =  Tt⁶ / 6.4 × 10¹⁹

Tt⁶ = 68.7484 × 6.4 × 10¹⁹

Tt⁶ = 4.3998976 × 10²¹

Tt = ⁶√(4.3998976 × 10²¹)

Tt = 4047.999 ≈ 4048 K

Therefore, the stagnation temperature Tt is 4048 K

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