Homework Statement
This should be a really easy circuit, but I need to see if I have it done right.
1V source, 1A source and a 1 Ohm resistor in parallel (forgot to write the 1 Ohm resistor on the image.)
Find the power of each element.
http://img403.imageshack.us/f/circuit1.jpg/...
I'm still not sure that I know what I'm doing, but sticking to two loops helped; I have the correct answers now. This is such a headache, you guy have no idea how long Ive been stuck on that simple little problem, onto the next one.
Thanks for the help.
Homework Statement
Sorry for edit, accidentally posted.
Circuit is show in my attachment. I have 3 loops with three unknown currents.
8 Ohms of resistance down the left side.
6 Ohms of resistance and a 4V battery down the center.
4 Ohms of resistance and a 12V battery down the right side.
I...
http://www.youtube.com/watch?v=oM0oBuhTLRI[/youtube]
http://www.youtube.com/watch?v=oM0oBuhTLRI"
As some of you may be aware, the 150th anniversary of Charles Darwin’s Origin of Species is later this month, the 24th to be precise. In one week from today, a special abridged edition of the book...
Actually, looking back over the problem, I think I assumed my 2 X 2 cross product was a cofactor expansion, where it should have been a constant. This would have allowed me to simplify that result and get the 128 \pi^3 I needed in the numerator.
Cofactor expansion results in 0 \vec{i} +...
The original question is: Find the curvature of
\vec{r}(t) = 4cos(2 \pi t)\vec{i} + 4sin(2 \pi t)\vec{j}
I have a solutions guide and found the curvature using
K = \frac{||\vec{T}'(t)||}{||\vec{r}'(t)||}
but I wanted to try the cross product method to see if it is...
I've been using various methods of finding curvature, and using the forumla
K=\frac{||\vec{r}'X \vec{r}''||}{||\vec{r}'||^3}
I took the cross product of my two vectors and came up with (just assume this is correct, if my question isn't answered i'll post all related work)...
I thought this was interesting, it's from ZapperZ's thread in general discussion.
If you notice in the "Today's Top 20 Viral Videos" section just below, http://www.viralvideochart.com/youtube/sarah_palin_cant_name_a_newspaper_she_reads?id=xRkWebP2Q0Y" [Broken] is posted. She seems to have a...
This is a really awesome game. I wish I had the creativity of some of their forum veterans. Many of their members post solutions much like you guys are, check out a couple of solutions I found to level seven, I just made a long cart. =(
http://FantasticContraption.com/?designId=600855...
The sound is messed up on that one,
sounds much better IMO and a great promotion for the LHC. I found it as a featured article on Yahoo and since I first saw it almost a million others have viewed it.
So, here’s the situation. My roommate and I live on the third floor of our apartment complex. As you walk into the building and begin to climb the stairs to the third floor you would notice a tremendous increase in temperature regardless of the weather. We have air conditioning but its not...
I use This site which allows you to play around with latex, and when it’s output you can copy and paste the image into MSword or whatever you have.
Just another option.
I guess I phrased that poorly, I simply meant that the integral was in a specific form which can be dealt with easily when you look at the powers of Sin and Cos. In this case he would want to break up Sin^3(13x) using Sin^2(x) + Cos^2(x) = 1 because the power of the sine function is odd.
I don't think that you can simply use a u-substitution here, this appears to be a Trigonometric Integral to me, which has its own special rules.
Since this is in the form of Sin^n(u)Cos^m(u) where m and n are both integers. Where does your book tell you to start?
Hint: focus on m and n...
You should surely be able to see why using u-substitution with the natural log will not work. My guess is that you are just given that anything in that form is going to be arctan and you will learn how to do this later on with Trigonometric Substitutions.
Remember
Sin^2(x) + Cos^2(x) = 1...
You're right that you need to use integration by parts and you're going about it correctly as far as I can tell. You just need to take the integral of your v*du and add your constant.
If I were you, I’d start off by drawing a picture.
y^2 = 16x
isn’t very helpful in this instance, how could you get a better picture of the ground. Once you know what the ground looks like it becomes easier to see where the path of the watermelon intersects the cliff.
How would you...
Ok, I’m still learning this stuff too and apparently I don’t know it as well as I thought. Given this problem I would have solved it the same way, and came up with the same answer as technology.
The only equation I can find that that relates velocity and time (it’s not in the book that I...
I picked up Physics for Scientist and Engineers With Modern Physics along with a solutions guide for like $20 a few months ago.
It's two editions back from the book that's out now and was released in 2000, so far it seems to be really great. It also seems to follow the outline of the video...
Yes all the values converge to zero. I really don't like to use the squeeze theorem, whenever I have to use it it's just by trial and error.
There should however be another theorem which relates to sequences that jump back and fourth. It's called the Absolute Value Theorem, which would allow...
When trying to determine if a sequence converges it needs to approach a certain number. By pluging in numbers for the value of n you'll notice that
Cn = [(-1)^n * 1/n!]
will jump back and fourth between positive and negative values, and therefore isn't approaching a single value. You...
You have found the slope for any point of that function except where the slope is undefined or wherever the graph may cross itself. To find the slope you would simply plug in your x and y values. However as you said you want to know the horizontal tangents.
You can't just simply set the y...
You can also find the derivative if you use a u-sub rather than simplifying the natural log. You might notice that the 1 in the numerator is the derivative of the denominator.
Try \frac{u'}{u}