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JoeMK

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Posts posted by JoeMK

  1. Yes' date=' I´m pretty sure. You get an equation for the velocity which (directly taken from your post; didn´t check it) is v = sqrt(G*m1/r). Any change in m2 will not alter the value of v. You can also leave m2 in if it makes you feel better: v = sqrt[(G*m1*m2)/(m2*r)']. Still: Changing m2 doesn´t change v (replace m2 with 5*m2, for instance).

     

    The mass cancels out at a much more basic level, actually. The force excerted by a gravitational field g on a mass m is F = mg. Now from Newtonian mechanics, you know that F=ma => a = F/m. Hence, the acceleration of a body caused by a gravitational field is the same for all bodies, regardless of their mass: a = F/m = m*g/m = g.

    Two bodies which have the same initial positions and the same initial velocities at a given time and do always experience the same acceleration move exactly the same path. So if your heavier satelite starts at the same height and with the same initial velocity as the lighter one, it will have exactly the same orbit.

     

    EDIT: Yes, I know it sounds illogical (unintuitive fits better, here). But it´s also counter-intuitive that a feather falls down as fast as a hammer if there´s no friction due to air resistance.

     

    Thanks buddy

    It makes better sense now

  2. It seems illogical that a heavier object needs the same velocity to travel in orbit as a lighter object. I know that gravity acts on all bodies equally, but i'm really having troubles imagining it.

    using the formula:

    m2*(v^2)/r =G*m1*m2 / (r^2) assuming m2 is the mass of the satellite, it cancels out...

    are u sure that means that the mass of the body doesn't affect it's velocity

  3. There is a formula to calculate this

     

    F (centripetal) = m2*(v^2)/r (pulls you outwards)

    F (gravity) = G*m1*m2 / (r^2) (pulls you earthwards)

     

    m1 = earth mass

    m2 = spaceship mass (cancels' date=' look the formula, so is independent)

    v = spaceship velocity

    G = Constant

    r = distance to the earth center (radius)

     

    So when

     

    F(centripetal) = F (gravity)

     

    m2*(v^2)/r =G*m1*m2 / (r^2)

     

    The force you are receiving is 0.

     

    Using this formula you must calculate what speed you need to be in orbit, at that speed forces will cancel out and you will feel weightless and spaceship will keep at the same altitude (will not fall) .

     

    If you are using a "wrong" speed, one which does not make forces cancel out, you will start to feel the gravity and you will fall to earth.[/quote']

     

    you said that "F (centripetal) = m2*(v^2)/r (pulls you outwards)"

    Centripetal force pulls you inwards, then u must mean centrifugal force... however, u put the formula for centripetal, which must be wrong

  4. I know that the stars release large amounts of heat as they burn, but the temperature of the universe does not increase, as it is constantly expanding.

    If the universe was static, then it'll obviously be much hotter, and most probably we wouldn't be alive.

    I need help with the prediction, and as well as to connecting it with the idea of "Red Shifts" and "Cosmic Background Radiation"

     

    Thanks

  5. The absorption and re-radiation from solid and liquid objects

     

    We have to design our own comparison experiment, to find out how different substances absorb or reradiate heat(or light). We do not have to do the experiement.

    Suggested equipement: A heat or light source, a bar heater or an overhead projector, a heat detector, your choice of solids and liquids.

     

     

    Any ideas, just need ideas

  6. I'm not really sure what you want for the Force/acceleration graph. Both the force and the acceleration are constant in magnitude' date=' and, with respect to eachother, in direction. That is, the graph of the force vector (ie, the force in a given direction) will be a sine curve. The graph of the accelration vector (similarly) will be a sine curve. The same sine curve. So plotting one against the other just gets a straight line of F vs a (or, more properly, F/m vs a, if you want to take variable masses into account).

     

    [img']http://img340.imageshack.us/img340/2544/fmagraph8xu.jpg[/img]

     

    If you plot it with F vs a, then all you can find out is the mass, from F = ma. If you plot it as F/m vs a, then all you can find out is that F = ma.

     

    "Mass vs Distance" I'm utterly confused about, as it appears to have no meaning at all.

     

    massline2yh.jpg

     

    Unless you mean weight, in which case it'll be a hyperbola.

     

    Mass vs distance is kinda impossibble, cuz u cannot show any time or distance progression. so i agree on the graph

     

    I'm a bit unsure about the force vs acceleration graph however.

  7. damn good question!

     

    one I`de like to know the answer to also :)

     

     

    I suspect' date=' but don`t know, that it`s something to do with the molecules of each metal not being Bonded chemicaly and so are easy to move around with the application of heat, it`s ONLY a guess though! :)

     

    a bit like the difference between Dry sand and Wet sand in the wind, the wet sand has bonding energy (if that makes sense?).[/quote']

     

    It's because the impurities disrupt the chrystal lattices of the parent materials.

     

    So yeah YT you were basically right.

     

    Thanks Guys, much apreciated

  8. Ok, we got this other question

    Use the following information to design the experiment and so complete the questions applicable to it

    a. u are given a substance that you know is a soluble chloride (to be refered to as X.Cl.)

    b. X is a group 1 cation

    Perform a first hand investigation to measure and identify the mass ratios of metal to non metals in a compund and calculate the empirical formula

     

     

    Anyone got an idea to a procedure? ... i hope i ain't asking to much, thanks :D

  9. i wish we had infrared thermometers. but we're stuck with the standard themometers (300*c). Is there somekind of special technique we can use?The solder is in an evaporating basin, we're using a bunsen burner in a smoke cupboard. I know the melting point should be around 180, but we can't get that exact point when it melts, and there's gotta be some kind of technique to use.

  10. hey, just need some help here

    this for some school assignment. We made our own Solder by melting tin and lead in the correct ratio. Now we need to remelt the Solder to find its melting point. Look i now this might sound stupid, but what is the most accurate method of findng the melting point.

     

    this is very impotant, there are 2 questions about it that pop-up in this in this assignment:

     

    1> Suggest a suitable technique and sound experimental practise that should be used so the reasonably low melting point can be accurately measured and confirmed

    2> What changes did you make to the procedure in this experiment in an attempt to gain more accurate results. (It is very unlikely but possible that you used the correct procedure for finding the melting point of Solder)

     

    Ok i know these might sound like very simple questions, but that teacher wants us to synthesise on a very high level of thinking, so the answerers ain't quite as obvious as they may seem ...any help srongly apreciated

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