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anonymous

  • 4 years ago

use dimensional analysis to find a formula for the time it takes the marble to return to the surface of the earth

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  1. anonymous
    • 4 years ago
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    What marble? What's the setup?

  2. anonymous
    • 4 years ago
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    that's all it gave me

  3. anonymous
    • 4 years ago
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    If that was it, it would say "a" marble, not "the" marble.... was it thrown up? Just dropped?

  4. anonymous
    • 4 years ago
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    oh it does say "a" marble sorry, it says it is shoot up

  5. anonymous
    • 4 years ago
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    Okay so what do you think the time might depend on?

  6. anonymous
    • 4 years ago
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    What variables, I mean?

  7. anonymous
    • 4 years ago
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    seconds

  8. anonymous
    • 4 years ago
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    No, that's the unit that the answer will be in, but what variables would the time of flight depend on? Like maybe the mass of the marble, or the time of day you shoot it up.... :)

  9. anonymous
    • 4 years ago
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    well it does say but i could guess gravity?

  10. anonymous
    • 4 years ago
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    That would be a good guess, anything else?

  11. anonymous
    • 4 years ago
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    air resistance but i think this case we would neglect that

  12. anonymous
    • 4 years ago
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    True... one more thing, it's staring you in the face... :)

  13. anonymous
    • 4 years ago
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    the obvious ones are always so hard to think of... maybe the velocity that it is shot?

  14. anonymous
    • 4 years ago
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    There you go :) So let's say that \[T \propto g^x\cdot v^y \]

  15. anonymous
    • 4 years ago
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    What are the units on the left?

  16. anonymous
    • 4 years ago
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    time in seconds

  17. anonymous
    • 4 years ago
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    Great, now what about on the right? for example, \[g^\alpha\] has units of \[m^\alpha\cdot s^{-2\alpha} \]

  18. anonymous
    • 4 years ago
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    i'm so sorry for doing that, but please, can someone help me?

  19. anonymous
    • 4 years ago
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    mass, m/s and m/s^2 i think

  20. anonymous
    • 4 years ago
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    m is meters, a unit of length. The way to do it goes like this: I showed you the units that we get from the g. Including the velocity, we get this: \[ \text{units of } g^x v^y = \left(\frac{m}{s^2}\right)^x \cdot \left( \frac{m}{s}\right)^y = m^{x+y}{s^{-(2x+y)}}\]

  21. anonymous
    • 4 years ago
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    This must equal seconds to the power 1, so we can solve these two equations and figure out x and y. Since we have no unit of length, \[x+ y = 0 \rightarrow x = -y\] Plugging that in to the second equation and setting it equal to +1, \[-(2(-y) + y) = -(-y) = y = 1 \] So y = 1, x = -1, and apparently \[T \propto \frac{v}{g} \]

  22. anonymous
    • 4 years ago
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    As it turns out, the real answer is \[T = 2\frac{v}{g} \] but dimension analysis always gives us proportional relationship that doesn't take into account dimensionless factors like 2.

  23. anonymous
    • 4 years ago
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    okay makes sense thanks a lot, i know dimension analysis is supposed to make everything easier but it never clicked for me anyway thanks again

  24. anonymous
    • 4 years ago
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    No problem. It doesn't necessarily make problems easier, per se, but it provides a good way to check if your formula is correct, and it is a comparatively fast way to squeeze some information out of the problem without actually solving it.

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