- #1
fishnic
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1. What is the meaning, when a graph of Stopping Distance vs. Kinetic Energy is made, of the slope of the line? Justify this answer by showing how y=mx+b corresponds to the quantities you actually plotted.D is given in CM, and KE given in [itex]\frac{g\times cm^{2}}{s^{2}}[/itex]
Three points (From lab work):
D = 14.0cm, KE= 344,450[itex]\frac{g\times cm^{2}}{s^{2}}[/itex]
D = 40.0cm, KE= 911,250[itex]\frac{g\times cm^{2}}{s^{2}}[/itex]
D = 111.0cm, KE= 2,553,800[itex]\frac{g\times cm^{2}}{s^{2}}[/itex]
Related Equations
W = Fd*cosθ
KE = (.5)mv^2
KE(net) = W(net)*d
After doing a basic slope calculation of [itex]\frac{(y-y)}{(x-x)}[/itex], the numbers were inconsistent, and when dividing the units, I couldn't figure out what it represented.
I worked the slope's units out to be [itex]\frac{s^{2}}{g\times cm}[/itex]
I don't know where to go from here. Our physics teacher's answer to questions is that "[we] are accelerated students and should be able to find it out on our own" or that "[we] should see him after school" (which, he is never there). We JUST started working with these problems yesterday, and I want to understand this unit.
Three points (From lab work):
D = 14.0cm, KE= 344,450[itex]\frac{g\times cm^{2}}{s^{2}}[/itex]
D = 40.0cm, KE= 911,250[itex]\frac{g\times cm^{2}}{s^{2}}[/itex]
D = 111.0cm, KE= 2,553,800[itex]\frac{g\times cm^{2}}{s^{2}}[/itex]
Related Equations
W = Fd*cosθ
KE = (.5)mv^2
KE(net) = W(net)*d
After doing a basic slope calculation of [itex]\frac{(y-y)}{(x-x)}[/itex], the numbers were inconsistent, and when dividing the units, I couldn't figure out what it represented.
I worked the slope's units out to be [itex]\frac{s^{2}}{g\times cm}[/itex]
I don't know where to go from here. Our physics teacher's answer to questions is that "[we] are accelerated students and should be able to find it out on our own" or that "[we] should see him after school" (which, he is never there). We JUST started working with these problems yesterday, and I want to understand this unit.
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