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erin512
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Compare the equation for a line through the origin, y = mx, to the equation Fc=mv^2/r and explain what the slope of the graph of Fc vs. v^2 represents. Remember that the m in the equation for a line represents the slope, and the m in the centripetal force equation represents the mass of an object
 one year ago
 one year ago
erin512 Group Title
Compare the equation for a line through the origin, y = mx, to the equation Fc=mv^2/r and explain what the slope of the graph of Fc vs. v^2 represents. Remember that the m in the equation for a line represents the slope, and the m in the centripetal force equation represents the mass of an object
 one year ago
 one year ago

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IsTim Group TitleBest ResponseYou've already chosen the best response.0
Ouch my eyes. Is this centripetal acceleration or something? Or like Centrifugal force?
 one year ago

erin512 Group TitleBest ResponseYou've already chosen the best response.0
yes centripetal acceleration :)
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
In the Fc case, rewrite \[F_c=mv^2/r=v^2(m/r)\] What that says to me is because m is constant (ignoring relativistic effects) as v increases, r has to increase very rapidly to maintain a constant force because of the v^2 term in the numerator. If we aren't trying to hold the force constant, but instead hold the radius constant (as well as the mass), the force goes up quite rapidly and the shape of the curve is a parabola opening upward. If we plot F_c vs. v^2, (m/r) is just the slope of the curve. I can't say that I'm positive I've understood what the author of the question is after here...
 one year ago

erin512 Group TitleBest ResponseYou've already chosen the best response.0
wow you are a genius, thank you so much for your help once again @whpalmer4
 one year ago

erin512 Group TitleBest ResponseYou've already chosen the best response.0
can u possibly help me with just one more? i'll repost now, this one should be a bit clearer in its description
 one year ago

whpalmer4 Group TitleBest ResponseYou've already chosen the best response.1
I much prefer solving physics problems to this sort of thing!
 one year ago
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