Search results for query: *

Hey all, I was hoping someone would be able to point me at a good ordinary differential equations text. Basically right now I have been working out of 'ordinary differential equations" by boyce and brannon. but I am fed up with the book and really would just like a good book to be used as a...

Thanks for looking it over. Your right that is a much easier method. But in the second part of the question it states "show that when \gamma \geq 1 then the source must have an infinite lifetime. The way I took this is that if it is to have an infinite life then the \lim _{t...

If M(x,y) is in terms of both x and y then I would say no, because you only have one equation and two unknowns

Im sorry, I posted this in the wrong section, feel free to move it to the homework section. Hey guys, I've really been needing some help with this one. I am doing an assignment for Ordinary Differential Equations and I was hoping someone could help me out by looking over my work. I've been...

Right, and you know that mu_s and mu_k are related by the equation F(friction)=mu*(F(normal)) and because this is a frictionless, massless pulley we are able to relate the tensions of the two boxes.

Draw a free body diagram for each box/crate to see where the forces are acting.

Thanks, when I had googled a similar search nothing relevant came up but Ill use the search you used.

Does anyone know if there is a way to password protect your thumb drive. Or any other kind of security that I could add to it.

Arent there a ton of places where you can pay one flat fee to download unlimited music? I think they only charge if you put it on a portable device, I can't remember the exact details though.

Yeah I checked it twice, it says 3200 for the bit rate. I doubt that's the number we are looking for though. But I know I did it higher than 64kbs

Im not entirely sure, how do you find that out? Heres what I tried anyway. I opened the player then clicked on a song and clicked properties. Then clicked details and it said bit rate was about 3200kbps for most of the songs I checked. But that number seems really high to me.

You should check out the sandisk sansa. I have a 4gb version and I'm happy with it. Its basically an Ipod mini for a lot less money and it has a lot of features like a voice recorder and radio. I've used the recorder a couple of times in class and its pretty sensitive. You need to sit at the...

Its still too early to tell.

You should probably call their tech support because it sounds like something might not be working properly. Do you know if all the internal fans are running?

Hmm, well I am not really sure how you could change the temperature on a laptop. Are you using it on a soft material i.e. covering the exhaust vents like if it is being used on a bed?

Well firstly stop using that computer because at those temperatures you will wear out the motherboard or cpu really fast. In order to help you though we will need more information about your system. What cpu do you have and what model motherboard. Also list any other components in the case.

They do give mu as .3. I think I might be able to solve the problem now. If I can't Ill post back. thanks for the help.

That makes sense if the initial movement is vertical. But how could I find the tension if the initial movement is horizontal. Would'nt that bring us back to the 3 variable 2 equation thing? If it would help I could scan the problem from the book also.

The problem I am having is that in all the figures (in the image) I have 3 variables with only 2 equations. In figure 3 The only way I could solve for the variables would be to set the normal force equal to the weight, however if the tension is also pulling up it would decrease the normal...

The sum is the addition of the i, j, and k components of the vector. To find the magnitude of the sum you first need to add the components then find the magnitude of the resultant vector. i.e. take the square root of the sum of the squares.

I actually got to use one the other day at the i store. I didnt care for it, but it was at least an interesting thing. The touchscreen is actually a lot better then I had expected but I still had some trouble with hitting multiple letters on the keyboard.(I don't think it would be possible to...

A mixture of different energy sources would be the best solution. So as not to rely on only one thing.

Ill try and remember that next time I buy pens, though it will probably be a while since I use pencils 90% of the time.

Ive been using the bic atlantis pens and like them alot. They are pretty cheap too, which is always nice.

Born Texas. Living in New Mexico

Yes it is solvable. You have 2 eqns and 2 unkowns.

Unless I am missing something I thought the half angle formula was \tan \frac{\theta}{2} = \frac{\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}} Thats not the same as \tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta} right? Or am I making a mistake?

Almost, but not quite. It looks pretty tough to put it in that particular form. The way I mentioned works so I guess Ill just stick with that, especially since remembering all those identities is a pain. Thanks for the suggestions though.

I think I found it. It came to me when I went to get the mail. :) basically I have this tan(30)=\frac {\sqrt {1 - \cos^{2} \theta}}{1+\cos(\theta)} From here I just put it in a quadratic form and solved.

tan(30)=(sin(theta))/(1+cos(theta)) The only way I can solve this is by using the graph on the calculator. There must be a way to solve it by hand though but I can't find it. Maybe I am just not thinking straight but its really getting to me.

My power comes from a surge protector strip, so I guess that would throw out the dirty power idea. They RMA'd the board but I am still not sure what caused the problem. Hopefully when I get the new board it will say what was wrong with the old one.

Thats what I had always thought. Maybe the professor was thinking that it would be easier to solve quadratics since the ti-89 will solve them with a lot less effort. But this would'nt merrit spending 100 + dollars to simplify an already simple process. Also, Maybe to make vectors easier to work...

Thats a good point, I think Ill just wait and see what happens. He said that test problems would be of the same difficulty as the homework, so if the homework requires a better calculator I guess Ill know. The funny thing is, is that my school requires a graphing calculator, but you can't...

My professor today told the class(engineering statics) that we should invest in the ti-89 or an equivelant. He claimed that the ti-83 was good for middle school but not for University work. However, I was forced to buy a ti-83 because the math department at my school doesn't allow anything...

One year ago I would have voted AMD, but the intel core 2's beat anything AMD can offer right now. So I guess you could say I prefer whoever offers the better product.

Do you have dust problems doing that?

Ok Ill concede. But I was just trying to avoid directional confusion. Since the problem starts with an - direction for acceleration, and to simplify it I did the problem assuming a positive acceleration. Edit* there it should be fixed.

What do you mean there is no deceleration? If the force applied produces a larger acceleration than is present. Then that means in effect the force of friction must be slowing the object down or decelerating it. Is there a better way you could put it? Sometimes I'm a little confusing.

Well, not quite, the force applied is not the same as the force of friction because there is an acceleration. The only time the force of friction is the same as the force applied is when acceleration is zero. The reason they give you an initial force is so you can find the -acceleration due...

Well here is a neat link on the forces at work on a banked curve:http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html But it does include friction but it will at least give you a really good idea as to what exactly is at play here. Hope it helps.

I agree with Warren on this one. Overclocking voids most warranties on your computer. Which could be extremely expensive to replace. I've gone through 2 motherboards without making any modifications to the system and can say I would never take the risk in overclocking unless you have an...

Thats basically what I have concluded too. Thanks for all the help.

If I had money I would probably just update my system and not worry about it but I have to stick with what I have for now. I actually have'nt made any modifications to the board or to the bios. (save changing the power supply to cool'n quiet mode.) What do you think about the fan...

Yeah I really like asus as a company especially since they already replaced the same board six months ago without a hassle, although I doubt they would let me get a different board but I guess I can ask since Ill be corresponding with them anyway. I've read on their forums that this is a fairly...

well what does it mean if I connected the power supply motherboard and cpu and absolutely nothing happens? Because I've done that much and there isn't any activity. Should it even do anything if I take the cpu out?

No there arent any led's on the front or back. But if it was bad ram or graphics etc shouldn't it at least post?

Well I though I would give it a shot posting here since no other computer or help forum seems to come up with any results. So here goes. About six months ago my motherboard on my desktop computer went dead suddenly. It was on one minute, I left it to eat dinner (no more than 15 min) I came...

Not quite,your equation says deltax=.5(v+vo)t so delta x is a change in x or xfinal - xinitial so runnerA:x-3mi=.5(8mi/h)t runner B:x-1mi=.5(5mi/h)t so you have two equations with two unknowns solve for t you should get the same value of t for each runner. Once you find t you can find the...

You should not have two values for time here. This isn't possible, how can 2 runners meet at different times if they are running towards each other. They should only meet once so you should have one value for t. Lets break this down. The problem gives you a distance east of the flagpole for...

No problem, if you have any other questions just ask.