I am also interested in these integrals. Some functions are just easier to work with in polar coordinates. If I remember correctly, the integral used to find area is derived my summing up areas of infinitesimal sectors of a circle. The formula for area is
A=\frac{1}{2} \int f(θ)^{2} dθ
Wow, how did I not see it? In formula 4
Then A_b=2A_a !
So now formula 1 can be written as E_a=4E_b
There is an error in the 4th relevant equation. It should be 2emf=l(E_a+E_b)=lE_c because the closed path only travels through 1 round bulb, not 2.
and the rest is algebra
On the interval [0,1], which is where we are evaluating the integral, are there any points of discontinuity for values of x?
If so, then the integral is improper, and you must replace that x-value with a variable, then take the limit as that variable approaches the x-value from either the...
We haven't talked about resistance or Ohm's law yet. We can only use Kirchhoff's node rule,
"In the steady state, the electron current entering a node in a circuit is equal to the electron current leaving the node."
We can also use "i=nAuE" and the Loop Rule, where voltage around a closed...
Is this your answer to part one? If so, consider that these values will make the integral diverge, not converge. Also, please show us your work.
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\int_{0}^{1} x^{-p} dx = ?
When I cancel factorials, I usually expand the factorial into maybe three or more factors, e.g. n!=n(n-1)(n-2)... and (n-1)!=(n-1)(n-2)...
Then it is clear that \frac{n!}{(n-1)!}=n
Homework Statement
When a single round bulb of a particular kind and two batteries are connected in series, 5* 10^18 electrons pass through the bulb every second. When a single long bulb of a particular kind and two batteries are connected in series, only 2.5*10^18 electrons pass through the...
*Light bulb*
Haha, thank you! I guess it may have been easier to use the parallel axis theorem from the get-go. I_cm would then be 2(M/12*L^2) and the total mass would be 3M.
Yes, my picture shows rotation parallel to the y-axis. In that case, isn't the moment of inertia for the rod aligned with the x-axis 1/3*M*L^2 ? Couldn't I model the other two as a particle, distance L/2 from the axis of rotation, as I suggested above? Then the moment of inertia for the...
I would like to take this question to the end. I understand that the moment of inertia for the rod aligned with the y-axis is 1/3*M*L^2. Couldn't I model the other two rods as a particle, distance L/2 from the axis of rotation, using the formula I = M*(L/2)^2 ?