Technician a says that the water pump is a centripetal pump. technician b says that centripetal force is the outward force that
1 answer:
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(a)
We can solve the different part of the problem by using Gauss theorem.
Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:
where q is the charge contained in the spherical surface, so
Solving for E(r), we find the expression of the field for r<a:
(b) 0
The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.
So, if we use Gauss theorem for the region a < r < b, we get
however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so
q' = + q - q = 0
And so we find
E(r) = 0
(c)
We can use again Gauss theorem:
(1)
where this time r > b (outside the shell), so the gaussian surface this time contained:
- the charge +q at the centre
- the inner surface, with a charge of -q
- the outer surface, with a charge of +q
So the net charge is
q' = +q -q +q = +q
And so solving (1) we find
which is identical to the expression of the field inside the shell.
(d)
We said that at r = a, a charge of -q is induced. The induced charge density will be
where is the area of the inner surface of the shell. Substituting
q = 5.00 C
a = 0.18 m
We find the induced charge density:
(e)
We said that at r = b, a charge of +q is induced. The induced charge density will be
where is the area of the outer surface of the shell. Substituting
q = 5.00 C
b = 0.46 m
We find the induced charge density:
This problem provides information about the pressure and temperature ideal gases are studied at. The answer to the questions are that all molecules have the same density, 2.43x10²⁵ mol/m³ and 2.43x10¹⁹ mol/cm³.
<h3>Idela gases</h3>In science, we can start studying gases with the concept of ideal gas, as they do not collide one to another and are assumed to be perfect spheres with no relevant interactions.
In such a way, one can conclude that the <u>number density of all ideal gasses at SATP is the same</u>, as they are assumed to be perfect spheres with equal volumes per molecule.
Moreover, when calculating the number of molecules per cubic meter, one must use the ideal gas equation as:
And plug in the numbers we are given:
Lastly, we can calculate the molecules per cubic centimeter by performing the following conversion:
Learn more about ideal gases: brainly.com/question/26450101