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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
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What marble? What's the setup?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's all it gave me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If that was it, it would say "a" marble, not "the" marble.... was it thrown up? Just dropped?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh it does say "a" marble sorry, it says it is shoot up

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay so what do you think the time might depend on?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What variables, I mean?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, 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
 4 years ago
Best ResponseYou've already chosen the best response.0well it does say but i could guess gravity?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That would be a good guess, anything else?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0air resistance but i think this case we would neglect that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0True... one more thing, it's staring you in the face... :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the obvious ones are always so hard to think of... maybe the velocity that it is shot?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There you go :) So let's say that \[T \propto g^x\cdot v^y \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What are the units on the left?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Great, now what about on the right? for example, \[g^\alpha\] has units of \[m^\alpha\cdot s^{2\alpha} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm so sorry for doing that, but please, can someone help me?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0mass, m/s and m/s^2 i think

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0m 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
 4 years ago
Best ResponseYou've already chosen the best response.0This 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
 4 years ago
Best ResponseYou've already chosen the best response.0As 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
 4 years ago
Best ResponseYou've already chosen the best response.0okay 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
 4 years ago
Best ResponseYou've already chosen the best response.0No 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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