e is the base of a natural log
Use the formula p(h)= 10.13e^-0.116h h=40
ln p = ln 10.13 + ln e^-4.64 = 2.32 - 4.64=-2.32
p= e^-2.32=0.1 newtons per sq cm = 0.99* 10^-2 atm (almost a complete vacuum) pressure at sea level 1 atm
Hey again,
I also have a second problem that I have a problem with because of a certain "e" I can't seem to use in my graphing calculator. The problem itself looks like this:
"Atmospheric pressure changes as you rise above Earth's surface. At an altitude of k kilometers, where 0 < h < 80, the pressure in newtons per square centimeter is approximately modeled by the function P(h) = 10.13e^-0.116h
(a) What is the approximate pressure at 40 kilometers above Earth?
(b) What if the approximate pressure on Earth's surface?
(c) Does atmospheric pressure increase or decrease as you rise above Earth's surface? The negative exponent tells you the number will be growing smaller."
My main problem is the mysterious "e" in the equation. I can't input it into my graphing calculator. I use 2nd > e on my TI-84 but that makes the problem come back as an error. Otherwise I might have a chance at solving the problem. Thanks!
e is the base of a natural log
Use the formula p(h)= 10.13e^-0.116h h=40
ln p = ln 10.13 + ln e^-4.64 = 2.32 - 4.64=-2.32
p= e^-2.32=0.1 newtons per sq cm = 0.99* 10^-2 atm (almost a complete vacuum) pressure at sea level 1 atm