atmospheric pressure. atmospheric pressure P (in kilopascals, kPa) at altitude h (in kilometers, km) is governed by the formula In (P/Po)=-h/k, where K=7 and Po=100 kPa constants. solve equation of P. use part (a) to find the pressure P at an altitude of 4 km
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OpenStudy (danjs):
hi
OpenStudy (anonymous):
hi
OpenStudy (anonymous):
dont know how to solve it
OpenStudy (danjs):
In (P/Po)=-h/k is given
OpenStudy (anonymous):
yes
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OpenStudy (danjs):
they also give you Po and k
OpenStudy (anonymous):
yes
OpenStudy (danjs):
so
ln(P/100) = -h / 7
OpenStudy (anonymous):
k=7 and Po=100 kPa
OpenStudy (danjs):
we want to solve this for P
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OpenStudy (anonymous):
ok
OpenStudy (anonymous):
yes
OpenStudy (danjs):
recall:
ln(a/b) = ln(a) - ln(b)
OpenStudy (anonymous):
yes i remember
OpenStudy (danjs):
ln(P) - ln(100) = -h / 7
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OpenStudy (danjs):
ln(p) = -h/7 + ln(100)
OpenStudy (anonymous):
you added In(100) to both sides?
OpenStudy (danjs):
yes
recall:
\[e ^{\ln(a)}=a\]
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
whats the next step
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OpenStudy (danjs):
ok
OpenStudy (danjs):
\[e ^{\ln(P)} = e ^{(-h/7 + \ln(100))}\]
OpenStudy (anonymous):
oh the e value
OpenStudy (danjs):
recall:
\[e ^{a+b} = e ^{a}*e ^{b}\]
OpenStudy (danjs):
\[P = e ^{-h/7} * e ^{\ln(100)}\]
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OpenStudy (danjs):
\[P = 100*e ^{-h/7} \]
OpenStudy (danjs):
For h = 4 km, What is P?
OpenStudy (danjs):
you get everything?
OpenStudy (anonymous):
oh sorry no
OpenStudy (anonymous):
Po=100 kPa
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OpenStudy (danjs):
Yes, we want to find P when h = 4 km
OpenStudy (danjs):
we solved for P in terms of h above
OpenStudy (anonymous):
oh for part A the book has e^-kh
OpenStudy (danjs):
\[\ln(\frac{ P }{ 100 })=-h/k\]
solving for P,
\[P = 100 * e ^{\frac{ -h }{ 7 }}\]
now you are given h = 4
OpenStudy (anonymous):
i need help with another one
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