A car is traveling 20 meters per second and is brought to rest by applying brakes over a period of 4 seconds. What is the averag
1 answer:
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The solution can be given using the Gauss' Theorem. First let's make the following assumptions:
*This problem has a perfect spherical symmetry so the field for a charge density of
that depends only on the radial component(we are of course working in spherical coordinates) will yield a constant electric field pointing radially outwards,that is, the field will be uniform for a<r<b independently of the angles 
and 
*As a result of the uniformity of the electric field, E will be constant and it can be taken out of the integral involving Gauss' law.
a)
Now let's get our hands to it, recall that Gauss' Law states:
Here we exploited the justified assumption that The field is uniform, because the field is pointing in the outward radial direction the scalar product between the field and surface element will yield
Now we need to determine the enclosed charge:
In spherical coordinates we thus have:
phi[/tex] over all of the gaussian surface.
Where the integral:
Returning to our integral we have:
![\frac{1}{\epsilon_0}\int^b_a\int^{2\pi}_0\int^{\pi}_0\frac{\alpha}{r}r^2sin\theta drd\theta d\phi=\frac{1}{\epsilon_0}\int^a_b\frac{\alpha}{r}r^2dr\int^{2\pi}_0\int^{\pi}_0sin\theta d\theta d\phi=\frac{1}{\epsilon_0}4\pi \int^b_a\alpha rdr \\ \\ =\frac{1}{\epsilon_0}4\pi\alpha\left[\frac{1}{2}r^2\right]^b_a=\frac{1}{\epsilon_0}2\pi\alpha(b^2-a^2)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Cepsilon_0%7D%5Cint%5Eb_a%5Cint%5E%7B2%5Cpi%7D_0%5Cint%5E%7B%5Cpi%7D_0%5Cfrac%7B%5Calpha%7D%7Br%7Dr%5E2sin%5Ctheta%20drd%5Ctheta%20d%5Cphi%3D%5Cfrac%7B1%7D%7B%5Cepsilon_0%7D%5Cint%5Ea_b%5Cfrac%7B%5Calpha%7D%7Br%7Dr%5E2dr%5Cint%5E%7B2%5Cpi%7D_0%5Cint%5E%7B%5Cpi%7D_0sin%5Ctheta%20d%5Ctheta%20d%5Cphi%3D%5Cfrac%7B1%7D%7B%5Cepsilon_0%7D4%5Cpi%20%5Cint%5Eb_a%5Calpha%20rdr%0A%5C%5C%0A%5C%5C%0A%3D%5Cfrac%7B1%7D%7B%5Cepsilon_0%7D4%5Cpi%5Calpha%5Cleft%5B%5Cfrac%7B1%7D%7B2%7Dr%5E2%5Cright%5D%5Eb_a%3D%5Cfrac%7B1%7D%7B%5Cepsilon_0%7D2%5Cpi%5Calpha%28b%5E2-a%5E2%29)
Now it's just a matter of solving for E:
b)
If a point charge is placed at the center of our system the the resulting field will be the sum of both fields, this field needs to be constant, let's pick for now that the field is zero in the region a<r<b:
Where
is the field due to a point charge.
Again using Gauss's theorem the field of a point charge 1 is:
We then get the following expression:
Solving for q we get:
If instead we want a non zero field
then we only need to solve
which yields:
![E_p+E=c \\ \implies \frac{\alpha}{2\epsilon_0}\frac{b^2-a^2}{r^2} +\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}=c \\ \\ \implies q=-2\pi[\alpha(b^2-a^2)-c]](https://tex.z-dn.net/?f=E_p%2BE%3Dc%20%5C%5C%20%5Cimplies%20%5Cfrac%7B%5Calpha%7D%7B2%5Cepsilon_0%7D%5Cfrac%7Bb%5E2-a%5E2%7D%7Br%5E2%7D%20%2B%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7Bq%7D%7Br%5E2%7D%3Dc%0A%5C%5C%0A%5C%5C%0A%5Cimplies%20q%3D-2%5Cpi%5B%5Calpha%28b%5E2-a%5E2%29-c%5D)
*This problem has a perfect spherical symmetry so the field for a charge density of
*As a result of the uniformity of the electric field, E will be constant and it can be taken out of the integral involving Gauss' law.
a)
Now let's get our hands to it, recall that Gauss' Law states:
Here we exploited the justified assumption that The field is uniform, because the field is pointing in the outward radial direction the scalar product between the field and surface element will yield
Now we need to determine the enclosed charge:
In spherical coordinates we thus have:
Where the integral:
Returning to our integral we have:
Now it's just a matter of solving for E:
b)
If a point charge is placed at the center of our system the the resulting field will be the sum of both fields, this field needs to be constant, let's pick for now that the field is zero in the region a<r<b:
Where
Again using Gauss's theorem the field of a point charge 1 is:
We then get the following expression:
Solving for q we get:
If instead we want a non zero field
Answer:
A) ≥ 325Kpa
B) ( 265 < Pe < 325 ) Kpa
C) (94 < Pe < 265 )Kpa
D) Pe < 94 Kpa
Explanation:
Given data :
A large Tank : Pressures are at 400kPa and 450 K
Throat area = 4cm^2 , exit area = 5cm^2
<u>a) Determine the range of back pressures that the flow will be entirely subsonic</u>
The range of flow of back pressures that will make the flow entirely subsonic
will be ≥ 325Kpa
attached below is the detailed solution
<u>B) Have a shock wave</u>
The range of back pressures for there to be shock wave inside the nozzle
= ( 265 < Pe < 325 ) Kpa
attached below is a detailed solution
C) Have oblique shocks outside the exit
= (94 < Pe < 265 )Kpa
D) Have supersonic expansion waves outside the exit
= Pe < 94 Kpa
