Unit for the slope of a circular motion relationship

[Epsillon]

Homework Statement

so given the formula T=(√((4π^2 m)/Mg)) √l+0
which is y=mx+b m being (√((4π^2 m)/Mg)) and x being √l

what is the unit of the slope??

The Attempt at a Solution

So the kgs cross out in the mass and we are just left with the units of g which is m/s^2
since g is on the denominator i got the unit s^2/m

HOWEVER my problem is that when the slope is multiplied by the x shoudnt it give you the units for y?

The way I did it it doesent.

 

[tiny-tim]

Hi Epsillon!

… and we are just left with the units of g which is m/s^2
since g is on the denominator i got the unit s^2/m

erm … it's √g.

 

[baba89]

The unit for the slope of a circular motion relationship is not just s^2/m, as it only takes into account the units of acceleration. The slope of a linear relationship represents the change in the dependent variable (y) for every unit change in the independent variable (x). In this case, the slope represents the change in time (T) for every unit change in the square root of the length (sqrt(l)). Therefore, the unit for the slope would be s/m^0.5, as it takes into account both the units of time and length. When multiplied by the units of the independent variable, the resulting units would be the same as the dependent variable (time in this case). So the units for y would still be seconds (s).