How much did the pressure drop in the storm's center from November 9, 1200z, until November 11, 0000z
1 answer:
The pressure dropped by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.
Using the attached map below, in the morning of November 9, the pressure situated at the storm's center = 1000 MB isobar located at the center.
Meanwhile, on November 11, 0000z the pressure situated at the storm's center = 976 MB isobar
The difference in this pressure is regarded as the pressure drop in the storm's center and it is determined as follows;
= 1000 MB - 976 MB
= 24 MB
Therefore, we can conclude that the pressure drop by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.
Learn more about pressure drop in an isobaric process here:
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Answer:
110.9 m/s²
Explanation:
Given:
Distance of the tack from the rotational axis (r) = 37.7 cm
Constant rate of rotation (N) = 2.73 revolutions per second
Now, we know that,
1 revolution = radians
So, 2.73 revolutions =
Therefore, the angular velocity of the tack is,
Now, radial acceleration of the tack is given as:
Plug in the given values and solve for . This gives,
Therefore, the radial acceleration of the tack is 110.9 m/s².