Help With Graphs Homework: Units of Slope, Slope Value, & Percent Error

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The discussion focuses on determining the units of slope for a graph depicting the relationship between the period and weight of washers. The slope is clarified to have units of (1/t^2)/N, indicating a relationship between the variables. Dimensional analysis is emphasized to ensure that the units on both sides of the equation align correctly. A calculation of percent error is provided, yielding a result of 15.02%. The conversation concludes with confirmation of the correct units for the slope, reinforcing the understanding of the relationship between the variables in the graph.
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Homework Statement


I need to know the units of the slope. What the slope should be. And the percent error.
Pictures:
http://imageups.com/files/101/456.PNG

Graph 1 is function of period(1/s^2) v weight of washers(N)
r=.77m
mass of rubber stopper= .02kg

Homework Equations



f(t)=1/t^2
w=mg
1/4pie^2(mass of rubber stopper)r

The Attempt at a Solution



would the slope be m/s*kg? or (1/s^2)/N?

I did 1/4pie^2(.02)(.77)= 1.644
(1.644-1.892/1.644)x100=15.02%

I think that is right
 
Last edited:
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Are you sure the abscissa has units of 1/s^2? You said period, which implies units of 1/time, not 1/time2.Dimensional analysis to the rescue:
You have a line, y=mx+b. However, these are not just numbers. They are things with dimensions. So, apply the units operator to the equation of a line:

units(y) = units(m)*units(x) + units(b)

Each term on the right must agree with the left-hand side:

units(b) = units(y)
units(m)*units(x) = units(y)

The first simply says that the intercept has the same units as does y. The second says that the slope has units equal to units(y)/units(x).

Example: Suppose the graph is of position in meters (abscissa) versus time in seconds (ordinate). Thus in this example, the y-intercept has units of meters and the slope has units of meters/second.
 
Yeah its suppose to be 1/t^2. I got that from Ac=(4pie^2r)/(t^2)
(4pie^2r) is constant in the lab so it would be one. I had to do this since the graph that was just period v weight wasn't a straight line. And we had to find a way to get a straight line.

1/t^2=(1/t^2)*Nx+1/t^2

so it would be (1/t^2)*N?
 
Last edited:
Correct.
 
Thanks :)
 
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