Question

The atmospheric pressure p decreases exponentially with the height h (in kilometers) above the Earth according to the function p equals 101.1 space e to the power of bevelled fraction numerator negative h over denominator 6.4 end fraction end exponent (kPa). Determine the height at which the atmospheric pressure is 25 kPa.

Answer 1

The atmospheric pressure is 25 kPa at a height of 8.95 kilometers above the Earth's surface.

Explanation:

To determine the height at which the atmospheric pressure is 25 kPa, we can use the given exponential function:

p = 101.1 * e^(-h/6.4)

We want to find the value of h when p = 25. Let's substitute the values into the equation and solve for h:

25 = 101.1 * e^(-h/6.4)

Divide both sides of the equation by 101.1:

25/101.1 = e^(-h/6.4)

Now, take the natural logarithm (ln) of both sides to eliminate the exponential:

ln(25/101.1) = -h/6.4

To solve for h, multiply both sides by -6.4:

h = -6.4 * ln(25/101.1)

Using a calculator, we can evaluate the right-hand side of the equation:

h ≈ -6.4 * ln(0.247)

h ≈ -6.4 * (-1.3989)

h ≈ 8.95 km

Therefore, the atmospheric pressure is approximately 25 kPa at a height of 8.95 kilometers above the Earth's surface.