5614

... and sometimes I'd use vf and vi for final and initial velocity. For the equations I mentioned above we were supposed to be taught to use x instead of s, for distance, in our physics A Levels, but nearly all us students were used to s from maths mechanics, so we continued to use s. I must say it is annoying having all this different notation for the same thing. In A Levels there were rarely minor differences between different subjects, but now I've come to university there's a lot of different symbols, and all us students, coming from different places and countries, are all used to different symbols from each other and the course! You said E=U/l, I'd say E=V/d etc. Our quantum course uses [math]\psi(x,t)=a \cos(kx-\omega t + \phi)[/math], whilst our vibrations and wave course uses [math]\psi(x,t)=a \cos(\omega t -kx+ \phi)[/math]... how irritating!

Use: [math]s = ut + \tfrac{1}{2} a t^2[/math] s = distance travelled u = initial speed t = time a = acceleration and: [math]v^2 = u^2 + 2as[/math] v = final speed all other symbols are same as above. For the reaction time one: if it takes you 1sec to react, and you're travelling at 10m/s, then you'll travel 10m before you start braking. So treat it as a normal question, but then add 10m on to account for your reaction time. I made these numbers up, work them out for your question. 4) presumably you have come across the equation: [math]F=\frac{d(mv)}{dt}[/math] in some form or another. F is the force, d(mv) is the change in momentum (also sometimes written as dp, as p=mv), and dt is the change in time. So for example if something of mass 1kg, at an initial speed of 2m/s, brakes to 0m/s, in 1sec, then: d(mv) = 1kg*2m/s - 1kg*0m/s = 2, and dt = 1sec, and then the constant force that is applied to obtain that constant deceleration (or momentum change, it's the same thing) is d(mv)/dt = 2/1 = 2N. Clearly I haven't given you the answers, but I think I've given you all the tools and equations you need, so now it's up to you to at least make a start. If you get stuck then post where you're up to and we'll help you out from there.

That's an unusual, although evidently not non-existant*, notation. What books do you have that use the notation (what subject and level are they)? * using a double negative made the sentence sound better, IMHO.

Think of it as water moving down pipes. In a series circuit all the water goes through all the pipes, it can't just dissapear. In parallel if there's a water pressure (equivalent to voltage) then most of the water will go down the path of least resistance, but a little bit of water will still go down the other paths, unless there's a complete blockage (i.e. infinite resistance).

U? V is normally used for voltage or potential difference. U is used for energy (potential or internal).

Not quite, because your battery is supplying some voltage, V, and it cannot just "be higher". If you have a power supply giving out 9V and you have a 9ohm resistor then I=1A, but if you change it for a 18ohm resistor then I=0.5A, the voltage won't change, the battery gives out a constant voltage (in simple cases it does anyway!). Usually the power supply won't change voltage when you change the circuit, so putting in more resistance will mean that the current is lower. Resistance does oppose the current. But you can also say that, in your series circuit, larger resistances will have a larger voltage drop across them.

In Brownian motion you see a particle move in loads of random different directions, as if it were continually being hit by loads of little things. What could provide all of these little collisions? Atoms.

Theory in science has a slightly different meaning to theory elsewhere. Your definition of theory is the "elsewhere" type because it includes beliefs, this is very much a non-scientific definition. See here for more detail: http://en.wikipedia.org/wiki/Theory

See some of the FAQs here: http://www.esrb.org/ratings/faq.jsp Also I think ratings kind of change over time. Something which would have been an 18+ ten years ago is probably now only 15+, that's society for you. Halo, being an 8 year old game, is probably affected by this. I don't really know much about SOCOM to comment on it specifically.

(because rockets burn fuel and thus have a varying mass - to explain the joke - just in case!)

Basically: p=mv is correct, for all "normal" cases. But when dealing with particles, or more specifically a light particle (aka a photon), which has no mass, you cannot say p=mv, becuase m=0. So instead a different formula, [math]p = \tfrac{h}{\lambda}[/math], is used. Where h is Planck's constant (just some number) and lambda is the wavelength of the light (or photon - same thing).

The potential energy it gains by dropping 200m is found by using: PE = mgh You know the mass, g is acceleration due to gravity (you'll use 10, or 9.8 or 9.81) and h is the distance it falls. Then you know that all of that potential energy went in to kinetic energy, and you know the formula: KE = ½mv². So you say that all the PE goes to KE, therefore: PE = KE mgh = ½mv² gh = ½v² 2gh = v² v = √(2gh) That should help. You understand now?

Well ya got the divide by lambda bit right!

Well if the pressure is staying the same then there is no heating due to pressurisation. As it's just air and air you don't need to worry about the specific heat capacities, unless you're taking in to account heat loss through the walls and to objects in the room etc. But in it's simplest case: if you have 1litre of air at 20C, and add another litre of air at 25C, then overall you get two litres of air at 22.5C. But if you're keeping the volume constant, then if you filled half your room with the warmer air, then let the two parts (hotter and colder) mix, then overall the temperature would again average at 22.5C. If you're dealing with just air (not different materials, not worrying about heat loss etc.) then it's just like doing a weighted average. So if have 1L of air at 20C, and displace 1/5 of that out and replace it with 1/5L of 25C air, then let it all mix and come to an equilibrium, then the temperature will be 21C.

ha, how totally pointless, but kinda funny!

Sadly I can't help much because I'm a physics student, but if I had a question along these lines then I know that my university has a careers department that I could go to for help, does your university not have a similar facility? I'm sure others will be able to help more than me.

If you have two crayons that are entangled and you accidentally drop one of them then it ruins the entanglement or correlation between the two of them. If two photons are entangled, but one of them interacts with something (for example hits a material, where it is absorbed and then re-emitted, this is what happens when light hits a surface, i.e. how we see) then the two photons will no longer be entangled.

[edit] this is long, but it's all words and very little maths, spend a little while reading it and I'm sure you'll understand it well! This is the diagram of the question: except that I have accidentally drawn 5 north + 3 east, not west, but if you understand how to do this then you will understand how to do all similar vector additions. What the question is asking is: if I go 5m north, and then 5m west, where do I end up, and what's the starting to end point distance and direction? As can clearly be seen in the diagram you start at the bottom and end up in the top right. Hopefully you have done Pythagoras' theorem (or Pythagorean theorem), and you can therefore work out the length of the blue line, which is what you get when you add the two black lines together (the details of the blue line is what the question is asking for). Pythag tells you that: length of blue line squared = length of the black line squared + the lenght of the other black line squared. Or in maths: (length of blue line)² = 5² + 3² length of blue line = √(5²+3²) = √(25+9) = √(34) = 5.8 As for the angle. Well you now have all 3 lenghts, although it is best to use the two you know exactly (the 5 and 3). Then you have to use trigonometry. Do you know SOH CAH TOA? That's what most people in the UK seem to use. In your question and in my diagram we want the bottom angle, and therefore the 5 and 3 lengths are the adjacent and opposite. As we have adj and opp, we want to use TOA, or tan. If you call the angle at the bottom of the triangle θ, then you get: tanθ = opp / adj = 3 / 5 therefore: θ = atan(3/5) = 31°. NB: atan = arctan = tan-1 (they're all different names for the same thing). Note you can click on two of the equal signs (square root and atan ones), this is where you would need a calculator. I hope this answers your question. Always draw a diagram!

*coughs innocently* *waves his hand in a jedi like fashion* "5614 knows the difference between red and green"

Dual boot with XP! When I got my laptop it came with Vista. So I immediately wiped it clean and then setup a XP/Vista dual boot, planning to use Vista as my main OS, but having XP as a backup. However my dislike of Vista very quickly grew and I now only use XP. One thing I did learn: finding XP drivers for my Vista-only laptop was kinda fun (and in the annoying sense of the word!). It seems to be a lot easier now, and there's a lot on the net about how to do it, but there wasn't as much when I did it, grr!

A really quick order of magnitude calculation I did (hopefully without a mistake, and just to keep my brain working!) shows that: given 700nm light, and 1kWm-2, then the impulse in that meter square, over 1 second was, per photon: 5*10-40Nm-1 and given that there's 4*1021 photons/sec in 1kW, impulse per m²: 2*10-18Nm-1. Saying the Earth is a disk of radius 6400km, gives a surface area facing the sun of: 1014m2. So a total force on the Earth due to the photons from the sun hitting it is about: 2*10-4N. However this doesn't account for the "counterbalance" of the Earth emitting photons. I was about to say that the dark and light side of the Earth balance each other out, but possibly the light side of the Earth would emit more photons, as it is in the light, thus increasing the force on the Earth due to all photon activity. However it should be a very small effect, even relative to the force I just calculated. [edit] wow, took me 15mins to type that, as Klay's post wasn't there when I stared! Klay: should be 1021 photons per kW, with both mine and your wavelength values. And about those wavelength values, I just chose 700nm as it's a little before green, and I was aiming for the middle of the visible light spectrum. Why did you chose 550nm, which looks like IR to me? Is sunlight mainly IR?

Almost:[math]p=\frac{h}{\lambda}=\frac{hf}{c}[/math] I know you know, just correcting your typo .

Ah! Thanks! WinISO can't see the extra file, but does see a boot information file. Whereas Altap Salamander (a similar program) sees an additional .img file, which is the same as WinISO's boot info file. I used WinISO to extract the boot information and can view/edit it with a hexeditor, which is what I wanted. I can then inject my modified boot file back in to the original .iso file, again using WinISO.

Here's a little puzzle I discovered and am working on, although I've run out of ideas! I've got a .iso file, and I want to see the content of it. When I burn it to a CD, load it as a virtual CD, explore or extract with WinRAR, I get this every time: there is 1.76MB of data, yet only a 2kb file is visible. I know there are other files within this .iso file, but I can't find a way to see them. Something which I just noticed is that the file system is CDFS. I've look this up a little bit, but don't fully understand it, and don't understand how it can hide files on a CD, nor how even knowing this if, and if yes then how, I can view the files on the CD. This isn't a trick question, the files aren't off the screen of anything! And yes I can see hidden and system files. Can anyone help?

The only relevant mention of multiverses in the orignal article I quoted in my last post, and no it does not state that. Beyond the quote, the author goes on to talk about how he likes the fact there might be another universe, exactly likes our, but with the gear stick in a Honda Accord a slightly different shade of grey... but nothing more relevant. The author was pointing it out as one of physic's more crazy theories, rather than actually detailing or discussing the theory. Indeed. This was my point.