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Chapter 15: Problem 28

Barometric pressure in the eye of a hurricane is 0.91 atm \((27.2\) in. of mercury). How does the level of the ocean surface under the eye compare with the level under a distant fair-weather region where the pressure is 1.0 atm?

Short Answer

Expert verified
The ocean level under the eye of the hurricane is approximately 93 cm higher compared to a distant fair-weather region.

Step by step solution

01

Understand Pressure Difference

The exercise provides two pressure levels, 0.91 atm in the eye of the hurricane and 1.0 atm in a distant fair-weather region. The difference in pressures is \(1.0 - 0.91 = 0.09\) atm.
02

Convert Pressure to Height

It is known that 1 atm of pressure corresponds to a 76 cm column of mercury. Using this, the pressure difference of 0.09 atm corresponds to \(0.09*76 = 6.84\) cm of mercury.
03

Convert to Water Level Difference

We also know that the density of mercury is 13.6 times the density of water. Therefore, the equivalent water height is \(6.84 * 13.6 = 93.02\) cm. This means that due to the lower pressure in the eye of the hurricane, the water level there would be approximately 93 cm higher than in the fair-weather region.

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Most popular questions from this chapter

A vertical tube \(1.0 \mathrm{cm}\) in diameter and open at the top contains \(\left.5.0 \mathrm{g} \text { of oil (density } 0.82 \mathrm{g} / \mathrm{cm}^{3}\right)\) floating on \(5.0 \mathrm{g}\) of water. Find the gauge pressure (a) at the oil-water interface and (b) at the bottom.

The pressure unit torr is defined as the pressure that will support a column of mercury \(1 \mathrm{mm}\) high. Meteorologists often give barometric pressure in inches of mercury, defined analogously. Express each of these in SI units. (Hint: Mercury's density is \(\left.1.36 \times 10^{4} \mathrm{kg} / \mathrm{m}^{3} .\right)\)

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