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:
brainly.com/question/13089696?referrer=searchResults
You might be interested in
Answer:
a
b
c
Explanation:
From the question we are told that
The mass of the bag is
The normal force experienced is
The maximum acceleration of the bag is
Generally this normal force experience by the bag is mathematically represented as
=>
=>
=>
=>
Generally for the bag not to slip , it means that the frictional force is equal to the sliding force
Hence is mathematically represented as
While is mathematically represented as
So
=>
=>
Generally from the workdone equation we have that
Here is the work done by friction which is mathematically represented as
Here s is the distance covered by the bag
is zero given that velocity at rest is zero
and
so
=>
substituting 2.55 m/s for v_i and 0.350 for \mu_k we have that
=>
Answer:
a) Fermi level = 600 electron-volts
b)
Explanation:
Given data:
length of one-dimensional crystal = 10 um
Lattice spacing = 0.1 nm
A) Determine the Fermi level assuming one electron per atom
Total length = 10 <em>u</em>m
Interatomic separation of a = 0.1 nm
in this case the Atom has one electron therefore the number of electrons = 10^5 and the number of states Ns = gsN = 2 * 10^5 ( attached below is some part of the solution )
hence : Fermi level = 600 electron-volts
B) Determine the density of states as a function of electron energy
attached below is the detailed solution
