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Everything posted by hotcommodity

  1. I'd pursue a degree in mathematics. It seems like the more math you know, the easier problem solving becomes.
  2. I know there are a few equations used to find the work done on an object, I'm just trying to use all of them to prove the same thing. But I'm still missing the negative sign when I use the magnitude definition of the dot product. Using vectors I'd have: [math]W_{grav} = \int^{h_f}_{h_0} \vec{F} \cdot d\vec{r} = \int^{h_f}_{h_0} <0, -mg, 0> \cdot <dx, dy, dz> = -mg \int^{h_f}_{h_0} dy[/math] Evaluate the integral: [math] W_{grav} = -mg(h_f - h_0) = -\Delta U [/math] Using components I'd have: [math]W_{grav} = \int^{h_f}_{h_0} \vec{F} \cdot d\vec{r} = \int^{x_f}_{x_0} F_x dx + \int^{y_f}_{y_0} F_y dy + \int^{z_f}_{z_0} F_z dz [/math] Zero force occurs in the x and z directions, so I get: [math]W_{grav} = -mg \int^{h_f}_{h_0} dy = -mg(h_f - h_0) = -\Delta U [/math] But when I use the definition of the dot product using magnitudes, I get: [math]W_{grav} = \int^{h_f}_{h_0} \vec{F} \cdot d\vec{r} = \int^{h_f}_{h_0} |mg||dr| cos\theta = mg \int^{h_f}_{h_0} dr = mg(h_f - h_0) = \Delta U [/math] The negative is missing in this case because I only use magnitudes. I'm not sure where my reasoning is flawed.
  3. I'd be integrating from the initial height to the final height.
  4. I see. But there's no way to use that definition of the dot product to find that its negative change in potential energy? I was told that some work definitions are for special cases. Edit: btw, I missed an integral in the opening post, sorry.
  5. But as far as the integral goes, I'm confused where the negative comes in? Is that an inappropriate way of finding the work done?..
  6. 61% You got 13 of the 21 people correct, and you did better recognizing the virginity of guys. Overall, you guessed better than 64% of all test takers. Girls have a poker face when it comes to virginity >:|
  7. That would be the work done by gravity. The work done by any conservative force is always equal to the negative of the objects change in potential energy, right?
  8. If work is definied as: [math]W_F = \int \vec{F} \cdot d\vec{r} [/math] And the dot product is: [math] |F||dr| cos(\theta) [/math] then I have [math]W_F = \int \vec{F} \cdot d\vec{r} = \int |F||dr| cos\theta[/math] Let's say that I wanted to show that the work done by gravity on a falling object, from [math] h_0 [/math] to [math] h_f [/math] is [math] W_{grav} = - \Delta U [/math] by using the integral of the dot product above. The displacement vector is in the same direction as the force, so [math] cos\theta = 1 [/math]. Now I have the integral of the magnitude of F, times the magnitude of dr. How do you end up with a negative in front of the change in potential energy when you have two magnitudes being multiplied? Any help is appreciated.
  9. I'm not sure what you mean by "motion for the 2nd..." If you're talking about motion in general, you could do a presentation on the x-prize. http://www.xprize.org/
  10. You can't really expect any institution to publicly back what Dr. Watson is saying, no institution is interested in having protestors at their doors. Chances are he wouldn't have declared this theory if there was any scientific argument that could readily challenge his belief. If he's not given a chance to explain his theory any time soon, it will probably take longer to debunk.
  11. Thank you for the reply. Ok, I get that it won't reach an absolute maximum, but if I consider "x" to be the distance that the spring compresses due to the block, can I think of that point as an end point graphically? In other words, can I plug in y = -[(.75m + x) sin(30°)] for the velocity function to obtain the maximum speed?
  12. Thank you for the reply. If I define the downward slope of the incline to be the positive x-axis, then the block will have a positive velocity. I defined h=0 to be the point at which the block starts accelerating, and [math]h_f[/math] to be the point at which the block begins decelerating. So I'd have: [math]0.5*m*v^{2}_f + mgh_f= 0[/math] [math]0.5*m*v^{2}_f = -mgh_f[/math] Now v final is a function of [math]h_f[/math] [math]v^{2}_f = \sqrt{-2gh_f}[/math] [math]\frac{dv_f}{dy} = \frac{-g}{\sqrt{-2gh_f}}[/math] Is this correct?
  13. You have a point, and I suppose I'm a little naive when it comes to the importance of voting. But lets say I research all of the candidates, follow all of the debates, and subsequently vote. How many voters would give that same consideration to each candidate in an effort to make an informed decision? I think alot of Americans, if not most, think of voting as nothing more than an American tradition, like the 4th of July. They label themselves "Republican," "conservative," "Democrat," and "liberal," and they let these titles, rather than information, decide their vote. I'm not very confident in the opinion of the masses. And I can't help but think that my vote would effectively work to counterbalance the votes cast by those who simply don't care to make an informed decision.
  14. Thank you for the reply. I'm still having a bit of trouble with this problem. If I understand what you're saying, the block will achieve its maximum speed right before it begins decelerating (and graphically, this makes sense), and it begins decelerating as it hits the spring. For part "a" of the problem, I found that the block does not compress the spring, x is equal to zero. So I think I only have to consider the movement of the block from the time it begins accelerating to the part right before it hits the spring. I'm not sure how to find this maximum speed, because in the dv/dy equation above, I have "g" in the numerator, and that's just a constant, it can never equal zero....
  15. I have a question on a homework problem having to do with a block accelerating toward a spring down an inclined plane. The problem is this: A 1.0 kg block starts at rest and slides a distance of 0.75 m down a frictionless 30 degree incline (with respect to the horizontal) where it hits a spring whose spring constant is 9.8 N/m. (No need to comment on my windows paint skills). Part "a" asks me to find the distance the spring is compressed, and I figured that out by using the the fact that the total mechanical energy of the system is conserved. Part "b" is what I'm having trouble with. It asks "where does the block achieve maximum speed during this slide." One of the hints my professor gave was to find a function of velocity and find its maximum by taking the derivative of that function. So this is what I did: I used the fact that mechanical energy is conserved from the time the block begins accelerating to right before it touches the spring. The energy terms below only deal with the block. [math]E_0 = E_f[/math] [math]U_f + K_f = U_0 + K_0[/math] The final potential energy, and the initial kinetic energy are zero if I take the point of the spring to be height = 0. [math]K_f = U_0[/math] [math]0.5*m*v^{2}_f = mgh[/math] [math]v_f = \sqrt{2*g*y}[/math] [math]\frac{dv_f}{dy} = \frac{g}{\sqrt{2*g*y}}[/math] It appears that it would achieve its maximum speed when the acceleration of the block is zero. If that's the case, this would be the point where the block is stopped by the spring. But if it's stopped, how can it have a maximum speed. Am I doing this wrong or am I missing a concept? Any help is appreciated.
  16. I've never voted before, but this year I've been thinking about it more seriously. Ron Paul caught my eye in the past year and a half or so, and I agree with alot of what he has to say. I doubt that I would vote for somebody that has alot of popularity behind them, so if I intend to vote for somebody less recognized, I have to ask myself : what are the chances of this person being elected? If they have a slim chance, I feel as tho' I'd be wasting my time casting a vote, as it would have little to no impact upon the results. On the national level, that's how I feel, but on the local level, I don't vote because I haven't had time to keep up with what's going on around me locally, and therefore, I feel as tho' my vote may cause more harm than good. As for people that simply do not care, as suggested in the opening post, it's probably better that they do not vote.
  17. How is it not his choice? Is somebody forcing him to be lazy? By the way, it's pretty common for an engineering degree to take nearly 5 years. All of the engineering graduates I know have taken 4 and a half to five years to finish up.
  18. His age shouldn't have anything to do with it. It doesn't matter how old you are, you can pursue what you want, as long as you mean business. I'm 24 and I'm a sophomore, but I don't let the age factor discourage me. One thing he'll have to get over is being lazy. You can't be an engineer and be lazy. At the very least it would cost him his job, and depending on the type of engineering he's pursuing, it could cost lives.
  19. Your friend sounds like a suitemate of mine who typically gets C's in his classes. But the thing is, when it comes to lab work and hands on stuff, he's top notch. And because of this, he already has a summer internship. Your friend would be better off pursuing what he's good at, what he has a knack for, and not worry so much about grades. Besides, he won't obtain an engineering degree unless he makes C's or better in all of his required classes. And if he's competent enough to do this, then I'm sure he can find something to do with his degree.
  20. Hmm...How is your teacher telling you to approach the problem with the KOH and H2SO4? If it's written down somewhere, type out the entire problem. This will give me a better grasp on what's being asked for.
  21. I assume your teacher is talking about titrating H2SO4 with KOH. Lets say you have plenty of KOH, but only a limited amount of of H2SO4, and lets say you don't know how much H2SO4 you're dealing with. The idea is to react just enough KOH with H2SO4. When this happens, your indicator in the titrated H2SO4 will change color. You want to look at how much KOH you used to titrate the H2SO4. Seeing as you have only a limited amount of H2SO4, you know that will be your limiting reagent. And by looking at the reation equation for KOH and H2SO4, you'll be able to determine the amount of H2SO4 you were initially given. Does this make sense?
  22. That's exactly what he was talking about. I watched the part of the lecture I missed online and my professor did a demonstration in class with the rope and the bicycle wheel, where it spins vertically without falling sideways due to gravity. I had to watch it 3 times to get a decent grasp on it, and it boggled my mind while I was trying to get to sleep. I took the test today tho', and all went well, I really do appreciate your help. I'll definitely check out the links.
  23. I see, I thought the torque vector would be along the axis of rotation, just as the angular momentum vector is, as you mentioned. Maybe I read the notes wrong, I'll have to go back and see. I remember a picture in my text book with the angular momentum pointing along the axis of rotation, the force pointing down, and the torque vector at a right angle to the force. I'll double check. I think I get what you're saying about the direction of vectors regarding f = ma. If I understand correctly, "a" is always a positive value, but as a vector a, it's a positive value "a" pointing in the negative i direction. That makes sense. Thank you for the help.
  24. I just have a few conceptual questions, and the first is regarding the direction of the torque vector of an object accelerating angularly. I was told in class that the torque vector points perpendicular to the vector of the angular momentum, and I have no idea why that is. There's lots of directions the torque vector can point and still be perpendicular to the vector of angular momentum, how can you tell? My professor mentioned something about the weight force of the object having something to with which way the rotating object will precess. I was late to class and only caught the tail end of the discussion. You always miss the good stuff when you're late. My second question is about vector notation. I'm a bit lost on how to write forces in vector form, as we tend to use magnitudes when we're adding up the forces that act on a body. When is it proper to write vector quantities in vector notation? For instance, if theres a frictional force decelerating a sliding box, and it's the net force acting on the body, how can I write that in vector form? I'm sorry if I appear a bit lost, but I'm trying to get what's going on here. So, we have the frictional force opposing the motion of the sliding box, and I know by looking at a free body diagram that it points in the negative x-direction, and I want to write the frictional force in vector notation. So, can I say f = ma, and a = -ai, and therefore the frictional force is -f ? I'm wondering if it would make sense to just place the negative in front of the force to begin with such as -f = ma, where a = -ai, but that would make f point in the positive direction. I know I've done it assbackwards a few times so I just want to get it right, any help is appreciated.
  25. If you'd like your thread to have more preciseness and specified direction, you have to put more thought into your opening post. You opened by saying: If you believe the people in this thread are getting off topic, you have no one to blame but yourself. You're trying to guide the thread after it's already begun, and you're attempting admonish people even tho' they've made no mistake. In short, what you'll get out of your thread is what you put into it.
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