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melamori03 [73]
3 years ago
9

How much did the pressure drop in the storm's center from November 9, 1200z, until November 11, 0000z

Physics
1 answer:
o-na [289]3 years ago
3 0

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:

brainly.com/question/13089696?referrer=searchResults

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What are the variables of Boyle's law?
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4 years ago
Five race cars speed toward the finish line at the Jasper County Speedway. The table lists each car’s speed in meters/second. If
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6 0
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Read 2 more answers
The tension in cable da has a magnitude of tda=6.27 lb. find the cartesian components of tension tda, which is directed from d t
Oliga [24]

Complete Question

The Complete Question is attached below

We have that the Cartesian components of tension T_{da} is

T_{DA}=-4.433i-3.49j+2.735k

From the Question we are told that

M_{da}=6.27 lb\\\\w=9.50ft\\\\d=6.60ft\\\\h=4.50ft

\vec {DA}=-4.7i-3.7j+2.9k)ft

\vec {DB}=-1.9i-3.7j+1.9k)ft\\\\\vec {DC}=-1.9i+5.8j-1.6k)ft

Generally the equation for T_{DA}  is mathematically given as

T_{DA}=\phi_{DA}* M_{da}

Where

\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{(-4.7)^2+(-3.7)^2+(2.9)^2}\\\\\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}

Therefore

T_{DA}=\phi_{DA}* M_{da}

T_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}* 6.27

T_{DA}=-4.433i-3.49j+2.735k

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