anonymous
  • anonymous
use dimensional analysis to find a formula for the time it takes the marble to return to the surface of the earth
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
What marble? What's the setup?
anonymous
  • anonymous
that's all it gave me
anonymous
  • anonymous
If that was it, it would say "a" marble, not "the" marble.... was it thrown up? Just dropped?

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anonymous
  • anonymous
oh it does say "a" marble sorry, it says it is shoot up
anonymous
  • anonymous
Okay so what do you think the time might depend on?
anonymous
  • anonymous
What variables, I mean?
anonymous
  • anonymous
seconds
anonymous
  • anonymous
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.... :)
anonymous
  • anonymous
well it does say but i could guess gravity?
anonymous
  • anonymous
That would be a good guess, anything else?
anonymous
  • anonymous
air resistance but i think this case we would neglect that
anonymous
  • anonymous
True... one more thing, it's staring you in the face... :)
anonymous
  • anonymous
the obvious ones are always so hard to think of... maybe the velocity that it is shot?
anonymous
  • anonymous
There you go :) So let's say that \[T \propto g^x\cdot v^y \]
anonymous
  • anonymous
What are the units on the left?
anonymous
  • anonymous
time in seconds
anonymous
  • anonymous
Great, now what about on the right? for example, \[g^\alpha\] has units of \[m^\alpha\cdot s^{-2\alpha} \]
anonymous
  • anonymous
i'm so sorry for doing that, but please, can someone help me?
anonymous
  • anonymous
mass, m/s and m/s^2 i think
anonymous
  • anonymous
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: \[ \text{units of } g^x v^y = \left(\frac{m}{s^2}\right)^x \cdot \left( \frac{m}{s}\right)^y = m^{x+y}{s^{-(2x+y)}}\]
anonymous
  • anonymous
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 \[T \propto \frac{v}{g} \]
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
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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