The atmospheric pressure P in pounds per square inch, at x miles above sea level is given by P=14.7e^-0.21x. At what height will the atmospheric pressure be one third of the sea-level pressure?
Thank you!!
Nevermind I got the answer, 5.73 miles.
Thank you!
p=14.63⋅0.8
a
To find the height at which the atmospheric pressure is one third of the sea-level pressure, we can set up the following equation:
P = 14.7e^(-0.21x)
1/3 * P = 14.7e^(-0.21x)
Let's solve this equation:
1/3 * P = 14.7e^(-0.21x)
Multiplying both sides by 3:
P = 44.1e^(-0.21x)
Now, we can set P equal to one third of the sea-level pressure, which is 14.7 (since 14.7 * (1/3) = 4.9):
44.1e^(-0.21x) = 4.9
Dividing both sides by 44.1:
e^(-0.21x) = 4.9 / 44.1
e^(-0.21x) = 0.1111
To solve for x, we can take the natural logarithm (ln) of both sides:
ln(e^(-0.21x)) = ln(0.1111)
-0.21x = ln(0.1111)
Dividing both sides by -0.21:
x = ln(0.1111) / -0.21
Using a calculator, we find that:
x ≈ -3.745
However, distance cannot be negative in this context, so the height cannot be negative. Therefore, we can ignore the negative sign and take the absolute value:
x ≈ 3.745
Thus, the height at which the atmospheric pressure will be one third of the sea-level pressure is approximately 3.745 miles above sea level.
To find the height at which the atmospheric pressure will be one third of the sea-level pressure, we need to set up the equation with the given information.
The sea-level pressure is represented by P0, which is equal to 14.7 pounds per square inch (psi).
We are looking for the height, x, at which the atmospheric pressure is one third of the sea-level pressure. This can be represented as (1/3)P0.
Using the given equation, which states that P = 14.7e^(-0.21x), we can set up an equation to solve for x:
(1/3)P0 = 14.7e^(-0.21x)
We can simplify this equation to:
(1/3)14.7 = 14.7e^(-0.21x)
Now, we can solve for x by isolating it on one side of the equation:
e^(-0.21x) = (1/3)
To remove the exponential term, we can take the natural logarithm (ln) of both sides of the equation:
ln(e^(-0.21x)) = ln(1/3)
Since the natural logarithm of e^(-0.21x) (where e is Euler's number) cancels out, we are left with:
-0.21x = ln(1/3)
We can solve for x by dividing both sides of the equation by -0.21:
x = ln(1/3) / -0.21
Using a calculator, we can find the value of x:
x ≈ -2.619
Therefore, the atmospheric pressure will be one third of the sea-level pressure at a height approximately 2.619 miles above sea level.