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uranmaximum [27]
2 years ago
5

Devise an exponential decay function that fits the given​ data, then answer the accompanying questions. Be sure to identify the

reference point ​(tequals ​0) and units of t. The pressure of a certain​ planet's atmosphere at sea level is approximately 800 millibars and decreases exponentially with elevation. At an elevation of 35 comma 000 ​ft, the pressure is​ one-third of the​ sea-level pressure. At what elevation is the pressure half of the​ sea-level pressure? At what elevation is it 2 ​% of the​ sea-level pressure?
Physics
1 answer:
7nadin3 [17]2 years ago
6 0

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

P=P_0e^{-kh}

where,

P_0 =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

P=\dfrac{1}{3}P_0

\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}

Now

P=\dfrac{1}{2}P_0

ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft

The altitude will be 22145.27733 ft

P=0.02P_0

0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft

The elevation is 124984.76055 ft

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