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Consider this scenario and answer the following questions: On a mid-August day in the northeastern United States, the following information appeared in the local newspaper: atmospheric pressure at sea level 29.97 in. Hg, 1013.9 mbar.

a) Convert 29.97 in. Hg to kPa.
b) Convert 1013.9 mbar to atm.
c) Compare the two pressure values.
d) Determine the significance of mid-August.

User Lumen
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Final answer:

The pressure of 29.97 in. Hg converts to approximately 101.5 kPa, and 1013.9 mbar to approximately 1.0006 atm. Both values represent standard atmospheric pressure. Mid-August signifies typical summer weather patterns and potential for variation in the northeastern United States.

Step-by-step explanation:

To convert atmospheric pressure from inches of mercury (in. Hg) to kilopascals (kPa), we exploit the fact that 1 in. Hg is approximately equal to 3.38639 kPa.

(a) The pressure in kPa is calculated as follows:

29.97 in. Hg × (3.38639 kPa / 1 in. Hg) ≈ 101.5 kPa

(b) To convert millibars (mbar) to atmospheres (atm), we use the relationship that 1 atm is approximately equal to 1013.25 mbar.

1013.9 mbar ÷ (1013.25 mbar / 1 atm) ≈ 1.0006 atm

(c) Comparing the two pressure values, we see that both indicate a pressure that is approximately equivalent to a standard atmospheric pressure at sea level, which is 101.325 kPa or 1 atm.

(d) The significance of mid-August in the context of atmospheric pressure may refer to local weather patterns during summer months. In many regions, mid-August is a time of stable, high pressure systems, although it is not uncommon for pressures to vary due to weather phenomena like hurricanes, especially in the northeastern United States. Meteorological changes in atmospheric pressure are especially noteworthy in forecasting storms and understanding seasonal climate patterns.

User Azheglov
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