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question 5 of 25
4 points
using the value of atmospheric pressure at sea level, $1 \times 10^5$ pa, estimate the total mass of the earths atmosphere above a $5-\text{m}^2$ area.
$9 \times 10^2 \text{kg}$
$4 \times 10^{-2} \text{kg}$
$5 \times 10^4 \text{kg}$
$2 \times 10^{-4} \text{kg}$
$3 \times 10^5 \text{kg}$

Explanation:

Step1: Relate pressure to force

Pressure \( P = \frac{F}{A} \), so \( F = P \times A \).
\( F = 1 \times 10^5 \, \text{Pa} \times 5 \, \text{m}^2 = 5 \times 10^5 \, \text{N} \)

Step2: Relate force to mass

Force \( F = mg \), so \( m = \frac{F}{g} \) (take \( g \approx 10 \, \text{m/s}^2 \)).
\( m = \frac{5 \times 10^5 \, \text{N}}{10 \, \text{m/s}^2} = 5 \times 10^4 \, \text{kg} \)

Answer:

C. \( 5 \times 10^4 \, \text{kg} \)