# Help with graphs

1. Oct 21, 2009

### kellenm

1. The problem statement, all variables and given/known data
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

2. Relevant equations

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

3. 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: Oct 21, 2009
2. Oct 21, 2009

### D H

Staff Emeritus
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.

3. Oct 21, 2009

### kellenm

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: Oct 21, 2009
4. Oct 21, 2009

### D H

Staff Emeritus
Correct.

5. Oct 21, 2009

Thanks :)