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Help with spring constant units
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[QUOTE="NewSoul, post: 4520081, member: 489477"] Ahh...it didn't mention in the instructions, but I just realized that g must mean the acceleration due to gravity. Gah! I knew it was something simple! With multiplying all of this by gravity, I end up with a much nicer 24.9 ± 0.10 kg/s[SUP]2[/SUP]. Thanks! Yes, you are correct. ------ Perhaps you can help me in calculating an error for part 2? Or do I have to make a new thread? Essentially... My equation is [B]k[B]d[/B] = (4π[SUP]2[/SUP])/S[/B]. (This is a new slope with s[SUP]2[/SUP] on the y-axis and g on the x-axis.) From the first equation, I found k[SUB]d[/SUB] = 24.1 kg/s[SUP]2[/SUP]. I also have an equation for a function [B]f(x) = Cx[SUP]n[/SUP], then... Δf/f = Cn(Δx/x)[/B]. Δf is the error, C is a constant, and Δx/x is the fractional error in x. Therefore, Δf must equal fCn(Δx/x). Now, it seems to me that this means Δk[SUB]d[/SUB] = k[SUB]d[/SUB](4π[SUP]2[/SUP])(-1)(ΔS/S). [I]However[/I], this would cause Δk[SUB]d[/SUB] to be -0.00190973753, which doesn't make sense to me. Shouldn't it be a positive number? So it seems then that k[SUB]d[/SUB] ± Δk[SUB]d[/SUB] = 24.1 ± 0.0 kg/s[SUP]2[/SUP]. [/QUOTE]
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Help with spring constant units
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