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mezarashi

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Everything posted by mezarashi

  1. Have you tried Ctrl+Alt+Del and closing any instances of it possibly trying ot run? And most importantly, explorer.exe. Shut that down. It causes alot of problems. Then go to Run-> explorer.exe to reboot it up. Of course, a restart would probably fix all of this as well.
  2. Thirteen is such a young age. I remember wanting to become an astronaut at that age, although my plans of obviously changed dramatically since ;P. There is alot more to science and engineering than just astronomy, but I guess that is what you are interested in at the moment. Strictly in the field, this can lead to graduate work in cosmology and astrophysics. For what you can do now is probably pick up as many books as you can in the library. Join clubs and activities that can enable you to learn. The astronomy club, electronics club, math club . I remember my astronomy club in high school had an optional astrophysics component organized by our club advisor that was reasonably challenging. Some high schools will also have an astrophysics 'introduction' in their courses. The problem with tackling these topics so early on is that you need mathematics before they can become enjoyable. The unforunate thing is that the mathematical difficulty increases exponentially once you enter college. But that doesn't stop you from becoming an expert at the simpler things like remembering every constellation for a particular day... atleast for now. There's quite a lot you can learn from just observing the night sky.
  3. The topic of heat transfer is a complicated one. Although it is often grouped together with a Thermodynamics course, the material itself can easily span a semester. You have to take into account the effect of not only conduction, but convection and radiation. Convection probably is what makes it most ugly. Going back to your problem, with the information you have provided, it is not possible to calculate the time. The specific heat only deals with "equilibrium" situations. Also with regards to the problem originally posted here, unless the problem states one can assume dQa/dt = dQb/dt, no information regarding the specific heat nor conductivity can be deduced from the temperature-time graph. If your "solution" is otherwise, then the question has silently assumed the condition of equal heat transfer over time.
  4. Firstly, your equation there is obviously not correct. C7H8 + 2KMnO4 → KC7H5O2 + 2MnO2 + KOH + H2O Working on the correct equation then, you can see that for each mole of toluene, you get one mole of potassium benzoate. However it says that you cannot expect more than a 71 percent yield. If my interpretation is correct, we can just alter this to be: for every mole of toluene, you can expect to get 0.71 moles of potatssium benzoate. From there, work using stoichiometry to find the molar mass of the two and solve ^^
  5. Your approach seems good enough ^^. Should be okay.
  6. Now that's very vague. I wouldn't know what to think '_';;
  7. There are few things that naturally travel at the speed of light. Electric fields and gravitational fields also propagate at the speed of light. Light can be considered a disturbance in the field and disturbances also travel at the speed of light.
  8. The only thing I can think of would probably be that if some parts of the other star reached into the event horizon of the black hole, that part would indeed be sucked in forever. Within this special radius, the black hole is official a black hole
  9. I would agree with that given that dQ/dt for both the materials are the same (they are dissipating the same amount of heat to the environment, or equally good at conducting heat to their surroundings), because dQ/dt is proportional to 1/mc x dT/dt. But there is no reason to assume so. So through the reasoning in my earlier post, I'd like to challenge this statement. I'd like proof to see through the heat transfer equations that indeed dT/dt not dependant on any other constants like thermal conductivity or diffusivity unless that can also be shown to be directly linked to a material's specific heat Cp, which would mean given a material's Cp we could determine its thermal conductivity and such. P.S.considering the time before freezing to be fair. Because one liquid happened to have a higher freezing point than another doesn't conclude anything about its specific heat.
  10. Your friends are right. When you study Thermodynamics, you study two things. Thermodynamics AND Heat Transfer. Thermodynamics and the concept of specific heat works without acknowledging the rate in which it happens. They usually assume enough time is given such that the system reaches equilibrium. Heat Transfer deals with the mechanism in which "heat" moves about. In a Temperature-time graph, you cannot conclude information on the specific heat. You can tell from the graph however that liquid A freezes at a higher temperature than liquid B. And you can tell that liquid B dissipates heat at a faster rate than liquid A. Similarly I can also tell that A takes a longer time to solidify than B. I can't conclude much regarding its specific heat or heats of fusion however. Kind of like chemical kinetics if you've studied that. Rates of reaction not related to Change in Internal energy. Liquid B may have had a high specific heat (high heat content), but it just gave away all that heat really really quickly. As for liquid A, it may have had a low specific heat (and thus low heat content), but it doesn't conduct its heat to air very well. A similar argument can be made for the freezing process. To test the specific heat of liquids. You need to have the two samples at the same temperature then put them into a volume of water and measure the equilibrium temperature of the water. The one with more heat content will definitely have more heat to pass on to the surrounding water thus raising its temperature more. And again, similarly for heats of fusion, you can melt the two substances in water and see the equilibrium temperature of water.
  11. Rust is hydrated iron(III) oxide. Water is a vital component to rusting. The water also helps remove the rust from the iron and thus you can have an iron bar "disappear" if exposed long enough. This is quite different from aluminum which has instead reacts to produce a "protective" aluminum oxide layer.
  12. I think that is a big overstatement. Even IF we do know all the rules of physics of the universe, it does not mean that we can harness it. Take for example we know that E = mc^2. There's ALOT of energy in mass, let's get it out then? But how? Apparently we are unable to. Do you think your college physics professor knows how to build a TV from scratch? A car? A computer? These are very low-tech tools by today's quantum standards. This is where we coin the term Technology. It is our ability to use our knowledge in science in useful ways. Simply, the applications. Alot of the patents flying around are based on techniques to get things done. They are no ground-breaking theories, just methods. I know this is generally unrelated to the thread, but I thought I'd correct some viewpoints
  13. If the walker in the in the middle, the sagging of the rope will make it form a triangle correct? The property of rope is that it can only support forces along its axis, that is it cannot exert torques. Resolve the horizontal and vertical directions. The two vertical forces must be equal to the weight of the walker. That's the key to solving this problem. Why don't you work out the problem and we'll see how we can assist you from there.
  14. Well yes, technically there's nothing wrong with that. However, you have to keep in mind that first of all, objects like the Earth are not composed of a singularity. That equation more or less is assumed to be true if the force originates from a point in space or a distance from a perfect sphere. If you've done your calculus exercises you will see that it cancels out to where a point mass estimate will be okay. Another factor to consider is that as r->0, there are 3 other forces (electromagnetic, weak, strong) that start to come into play which are MUCH MUCH MUCH stronger than gravity. Once you bring two atoms close to each other, their electron clouds (electromagnetic force which is about 10^39 times stronger than gravity) will interact repulsively.
  15. How about, the definition of average acceleration is simply: Final Velocity (can be a vector) - Initial Velocity (vector as well) over the time needed to accomplish this change in velocity.
  16. You have 3 variables and you have the three variables in an equation. What's wrong with it? You don't get resistance? But resistance is in the equation no? Elaborate. To solve for the proportionality constant, you would plug in the values of R, L, and d for a particular data point and solve for k. Given that your data is not perfect, your value for k will probably vary depending on the point that you use.
  17. Can you define what "statement" means? Unless I am misunderstanding you, I would believe you are looking for: R is proportional to L/d^2 ? which can be induced logically since we know that R is proportional to L L proportional to d^2 and R is proportional to 1/d^2 Your empirical equation would then be: R = kL/d^2, where k is some constant of proportionality.
  18. The folks at Sun Microsystems are probably going to be really really sad to hear that. They've been promoting their Java Standard Development Kit just about everywhere. I'm not sure if it comes with a development environment, but I think not (atleast not SDK). You'll need to run the compiler manually through a prompt or get a program (and there are many) that can automate this a bit for you. Oh, right. http://java.sun.com/
  19. Proportionality statement I see? This can be the easiest thing to do given any set of correlated data. I'm guessing maybe you don't fully understand what it means. They are basically asking you for the relationship between Length, Diameter, and Resistance. Take a look at your data. There are three variables. With one variable kept constant, increasing the second variable, how does the third variable change? Some hints, variables can be: - proportional to each other (momentum and velocity) - inversely proportional to each other (pressure and volume in ideal gases) - proportional to the square (kinetic energy and velocity) - inversely proportional to the square (gravitational force and distance between the masses)
  20. I see, so that's what it's called. Now that's an interesting phenomena I would say. Let's see if I can find more physics related to this tidal locking. Thanks for the pointer
  21. I don't see why the Earth-Moon rotation-orbit configuration should be preferred over any other. I think there needs to be an explanation as to why the Moon's "day" (one revolution around its own axis) is exactly equal to its "year" (one revolution around Earth). I find it very very odd for this to be the case. If the configuration is not preferred nor observed extensively in other orbital configurations, then the coincidence sets the ground for me to start believing in things like ID.
  22. I've read Michio Kaku's book Hyperspace, and personally I believe that it is very possible that the theory of hyperspace is true, although I find your comment "that hyperspace is a dimension" a bit confusing. The only problem here is that this continues to stay out of main stream science due to its untestability. The types of energy required to probe objects of Planck length like the supposed contracted dimensions in hyperspace theory or the strings in string theory are and will not be possible with our technological prowess anytime soon. Many theorists in the field try hard to find some observable measurement that can be attributed as a consequence of the theory. So far they haven't been very convincing. Hyperspace has many philosophical implications as well and its enigmatic nature is what makes even the general audience interested. Found some websites that show video of rotating hyper structures. The best way to understand these structures would be by observing the shadow of rotating a bare 3-D cube structure. How would you explain to the 2-D man what he is seeing? And what happens now if you rotate a 4-D cube and 5-D, etc... http://www.lboro.ac.uk/departments/ma/gallery/hyper/cube.html
  23. They do? Websites are usually not a problem because your side makes the initial request. I assume here that you are behind a firewall, such as a corporate or university firewall. I haven't seen the case of Windows' firewall causing such problems especially after SP2. There is only 1 solution I know of if you are indeed in such a predicament. There are many types of firewalls. Ones that block ports and allow only specific ports such as those you use to surf the web, or firewalls that sniff packets and block certain protocols for example FTP and such. More often however, you get the former. Being behind the firewall, with the only internet gateway being filtered by this very firewall, there is very little you can do to "bypass" it. You need to talk to your network administrator. The one solution that I have is to use a proxy server. IIRC, they are called High-anonymity proxies and will allow SSL sessions. The only catch is you may need to find one that uses an port that is open from behind the firewall. You basically enter this proxy address into your IE or firefox preferences.
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