Question
At a height of $10 \mathrm{~km}$ ( $33000 \mathrm{ft}$ ) above sea level, atmospheric pressure is about $210 \mathrm{~mm}$ of mercury. What is the net resultant normal force on a $600 \mathrm{~cm}^{2}$ window of an airplane flying at this height? Assume the pressure inside the plane is $760 \mathrm{~mm}$ of mercury. The density of mercury is $13600 \mathrm{~kg} / \mathrm{m}^{3}$.
Step 1
The pressure inside the plane is 760 mm of Hg and the pressure outside is 210 mm of Hg. So, the difference in pressure is given by: \[ P = P_{inside} - P_{outside} = 760 \, mmHg - 210 \, mmHg = 550 \, mmHg \] Show more…
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At a height of $10 \mathrm{~km}$ (33 $000 \mathrm{ft}$ ) above sea level, atmospheric pressure is about $210 \mathrm{~mm}$ of mercury. What is the net resultant normal force on a $600 \mathrm{~cm}^{2}$ window of an airplane flying at this height? Assume the pressure inside the plane is $760 \mathrm{~mm}$ of mercury. The density of mercury is $13600 \mathrm{~kg}$.
Normal atmospheric pressure is $1.013 \times 10^{5}$ Pa. The approach of a storm causes the height of a mercury barometer to drop by 20.0 $\mathrm{mm}$ from the normal height. What is the atmospheric pressure? (The density of mercury is $13.59 \mathrm{g} / \mathrm{cm}^{3} . )$
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