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 the sea level by the formula... p=760e^(-.145h)
Question. Find the height of a mountain if the atmospheric pressure is 664 millimeters of mercury.
2007-12-10 12:57:51 · 4 answers · asked by Dave & Sunny 2 in Science & Mathematics ➔ Mathematics
plug-n-chug: p = 664
664 = 760e^(-.145h)
332 / 380 = e^(-.145h)
83 / 95 = e^(-.145h)
.8736 = e^(-.145h)
ln .8736 = -.145h
-.1350 = -.145h
.9313 = h
The "mountain" (we have hills taller than that here in Ohio) rises about 930 meters above sea level.
2007-12-10 13:15:54 · answer #1 · answered by jimbob 6 · 0⤊ 0⤋
Assuming your forumla is correct... in what follows, "ln" refers to the "natural log". ("ln" is usually written in lower case, but the first letter is "L.") For a given number x, the natural log of x, written "ln(x)", is a value such that e ^ ln(x) = x. In other words, if x = e squared, ln(x) = 2. Many calculators have a "ln" key.
664 = 760 * e ^ (-.145*h)
664 / 760 = e ^ (-0.145 * h)
ln (664 / 760) = ln (e ^ (-0.145 * h))
ln (664 / 760) = -0.145 * h
(ln (664 / 760))/(-0.145) = h
Working this out, h is about 0.93 km.
2007-12-10 21:07:00 · answer #2 · answered by Anonymous · 0⤊ 0⤋
as p=664
ln(664/760)= -.145h
hence h= .9312 Km
2007-12-10 21:06:33 · answer #3 · answered by shrikant s 2 · 0⤊ 0⤋
p = 760e^(-.145h)
664 = 760e^(-.145h)
664/760 = 0.874 = e^(-.145h)
ln 0.874 = -.145h
(ln 0.874)/(-.145) = h
h = 931.1 m
2007-12-10 21:03:48 · answer #4 · answered by Philo 7 · 1⤊ 0⤋