The atmospheric pressure p on a balloon or an aircraft decreases with increasing height. This pressure, measured in millimeters of mercury, is related to the height h (in kilometers) above sea level by the formula p = 760e^(-0.145h).
(a) Find the height of an aircraft if the atmospheric pressure is 340 millimeters in mercury.
(b) Find the height of a mountain if the atmospheric pressure is 777 millimeters in mercury.
2007-12-29 08:34:32 · 5 answers · asked by journey 1 in Science & Mathematics ➔ Mathematics
Both problems are solved the same way. Substitute your known value for p, divide by 760 to isolate the e expression, take the natural log of both sides, and divide by (-0.145) to isolate h.
(a) Substituting for p,
340 = 760e^(-0.145h). Dividing by 760,
0.447368421 = e^(-0.145h). Take the log,
ln[0.447368421] = ln[e^(-0.145h)]
-0.80437281567 = -0.145h. Dividing by (-0.145),
5.54739872876 = h.
The height is about 5.5474 km.
(b) Substituting for p,
777 = 760e^(-0.145h). Dividing by 760,
1.022368421 = e^(-0.145h). Take the log,
ln[1.022368421] = ln[e^(-0.145h)]
0.022121917 = -0.145h. Dividing by (-0.145),
-0.15256494543 = h.
The height is about 0.15 km below sea level... not much of a mountain, if you ask me!
2007-12-29 08:55:26 · answer #1 · answered by Anonymous · 0⤊ 0⤋
a) set p=340 and you have:
340=760exp[-0.145h]
0.4474=exp[-0.145h]
take logs both sides:
-0.8044=-0.145h
h=5.54 km
b) the pressure here is greater than mean pressure of 760, so you will get an answer below zero
but, the answer you would get would be:
log[777/760]/(-0.145)=h
h=-0.15 kilometers (i.e., below sea level, please check the value you want for the pressure at mt top)
2007-12-29 08:49:45 · answer #2 · answered by kuiperbelt2003 7 · 0⤊ 0⤋
p = 760e^(-.145h)
Take natural logs
ln p = ln 760 -.145h
a) ln 340 = ln 760 -.145 h
5.8289 = 6.6333 -.145 h
h = 5.575 km
b) ln 777 = 6.6333 -.145 h
6.6554 = 6.6333 -.145 h
h = - 0.05 km (i.e., slightly below sea level)
2007-12-29 08:57:23 · answer #3 · answered by Joe L 5 · 0⤊ 0⤋
340 = 760 e^(-.145h)
17/38 = e^-.145h
ln(17/38) = -.145h
h = ln(17/38)/-.145 = 5.55 km
777 = 760 e^-.145h
h = ln(777/760)/-.145 = -.153 km
2007-12-29 08:47:16 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
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2016-10-20 07:45:30 · answer #5 · answered by ? 4 · 0⤊ 0⤋