Thanks for taking the time to answer my queries, I really appreciate it.
You are right, there are quite a few typos in the text. I guess my questions would be if:
$$ x^2 + x + 2 = (z-x)^2 $$ then: $$ \sqrt{x^2 + x + 2} = (z-x) $$ how did the author get: $$ \sqrt{x^2 + x + 2} = \frac {(z^2 + z...
Thanks, but my question is with regard to point 2 and 3. I am well aware about how to use the technique of point 1. Also, problem 1-3 are related to point 1.
Hi,
With respect to the techniques mentioned in point 2 and 3:
Can someone explain or even better, post a link for an explanation or a videos showing the use of these two techniques.
Below excerpt shows problems 4 and 5 referenced in the above 2 points:
Perhaps I wasn't clear. I understand how a straw works. My question is will fluid flow encounter more
energy losses in a small diameter pipe or in a larger diameter pipe. The typical answer normally given
is: for the same flowrate a larger diameter pipe will be more efficient (less energy...
Hello everybody,
It is normally said that fluid flow in a larger diameter pipe is easier (less energy losses) than in a smaller
diameter pipe. This indeed appears to be true in relatively high flowrate applications however, I'm struggling to
understand why drinking from a normal straw (small...
Thank for replying.
What I have understood from your reply is that when the inductor
is open-circuited it will have to spark (giving that no clamping method
is provided), and therefore all of my assumption (which is based on
the assumption that you could open-circuit an inductor without having...
Hello everybody,
First of all I want to thank you all for this great forum.
My question is if a superconductor inductor is charged with a certain voltage and then it is
open circuited what will happen.
My very basic understanding tells me that if the inductor is not a superconductor (it...