Respuesta :
20) We have to model the variation of the atmospheric pressure with the height.
We know that it is 1013 atm at sea level and it decreases by 12% for every kilometer climbed.
Then, we can consider the sea level to be h = 0.
We then will have a pressure P(0) = 1013.
We can also relate the pressures with one kilometer of difference as:
[tex]\begin{gathered} P(h+1)=P(h)-0.12P(h) \\ P(h+1)=(1-0.12)\cdot P(h) \\ P(h+1)=0.88\cdot P(h) \\ \frac{P(h+1)}{P(h)}=0.88 \end{gathered}[/tex]We can model this type of relations with an exponential model:
[tex]P(h)=A\cdot b^h[/tex]We then have to find the values of parameters A and b.
We then start with A, that will correspond to the initial value of P, as:
[tex]P(0)=A\cdot b^0=A=1013[/tex]Then, we can calculate b as:
[tex]\begin{gathered} \frac{P(h+1)}{P(h)}=0.88 \\ \frac{A\cdot b^{h+1}}{A\cdot b^h}=0.88 \\ b^{h+1-h}=0.88 \\ b^1=0.88 \end{gathered}[/tex]Then, b = 0.88 and we can write the model as:
[tex]P(h)=1013\cdot0.88^h[/tex]21) We can calculate the pressure when h = 5.895 km using the model as:
[tex]\begin{gathered} P(5.895)=1013\cdot0.88^{5.895} \\ P(5.895)\approx1013\cdot0.4707 \\ P(5.895)\approx477 \end{gathered}[/tex]22) We now have to calculate the pressure for the top of the Mt. Everest (h = 8.848 km):
[tex]\begin{gathered} P(8.848)=1013\cdot0.88^{8.848} \\ P(8.848)\approx1013\cdot0.3227 \\ P(8.848)\approx327 \end{gathered}[/tex]23) We finally have to calcualte the pressure for h = 0.383:
[tex]\begin{gathered} P(0.383)=1013\cdot0.88^{0.383} \\ P(0.383)\approx1013\cdot0.9522 \\ P(0.383)\approx965 \end{gathered}[/tex]Answer:
20) The model is P(h) = 1013*0.88
