How many grams of cobalt-60 will remain after three half-lives? A.2.50 g B.30 g C.1.25 g D.180 g
1 answer:
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Answer and Explanation:
1. Evaluate the function x(t) at t=0.5
2. The period of motion T can be calculated as:
Where:
So:
3. The angular frequency can be expressed as:
Solving for k:
4. Find the derivate of x(t):
Now, the sine function reach its maximum value at π/2 so:
Solving for t:
Evaluating v(t) for 0.6603981634:
So the maximum speed of the block is:
In the negative direction of x-axis
5. The force is given by:
The cosine function reach its maximum value at 2π so:
Solving for t:
Evaluating x(t) for 3.016592654:
Therefore the the maximum force on the block is:
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