Density (Linearized plot)
Billy and Mandy
In this lab the density of hand-made clay balls were calculated to understand how scientists model physical effects and to understand logarithmic plots. The hand-made balls ranged from diameters of 2cm to 6cm and were measured with vernier calipers by each member of the group. A total of 6 independent measures of each diameter were taken to establish uncertainty. The clay balls were then weighed on a gram slider and then recorded. The first data set was graphed mass (kg) against diameter^3 (m^3).
Yielding the linearized slope of density.
The density, ρ ±Δρ was calculated to be 1747+/- 11 (kg/m^3)
Assuming that the volume depends on diameter is unknown, volume was written: V=aD^n, where ‘a’ is a constant. The second data set was graphed lin(m) vs ln(D) in order to calculate the power ‘n’. The power ‘n’ was calculated to be:
n ± Δn= 3.3± 0.034 .
The objective of this lab was to determine the density of clay by weighing a set of hand-made spheres and model a function for mass vs. diameter using a linearized plot.
1. Make a range of clay balls whose diameters range from 1-7cm 2. Measure its diameter with vernier calipers
3. Take 6 independent readings to establish uncertainty
4. Weigh the sphere and wiggle the gram slider to estimate errors. 5. Repeat for the series of balls
Plot 1 table
Mass vs Diameter
m=1.093 +/- 0.01749
b=0.5972 +/- 2.107
Plot 2 Table
Ln(m) vs Ln(D)
m= 3.007 +/- 0.02603
b=-0.09895 +/- 0.03886
Lists the results of 6 different sets of diameter measurements: the spheres’ masses, the average diameters, the cubes of the average diameters, and the natural logarithms of average diameters and masses. These values were used to create two data plots.
The density of clay from the linearized plot of mass and the diameter cubed (m, D^3) data. Yielding a slope of 1.093 +/- 0.01749 m^3/g.
Since the density of an object is ρ=mV
And the formula for the volume of a sphere is V=π6D3
The density of a material shaped into a sphere would be equal to the volume of a sphere plugged into the density equation, which is:
Solved for D3: D3= 6πρm
In the linear plot, it is expectable to say the slope=6πρ
Using the value of slope, density can be solved for ρ=6π(slope) ρ=1747kg/m^3
The error of ρ(Δρ) can be calculated from the error of the slope: Δρ=|ρslope0+Δslope-ρslope0|
The density of the modeling clay with error is
1747+/- 11 kg/m^3
(This value seems reasonable when compared to the densities of other objects).
V=aDn, where ‘a’ is a constant. Ln(D) vs. ln(m) is plotted. The linear fit to this data provided a slope that could be used to calculate the power, n. The slope of the linear fit was:
This was solved by taking: m= ρV=ρaD^n
And then taking the natural log, ln, yields: ln(m)=ln(ρa)+nln(D)
Ln(D) is then solved for: ln(D)=1nlnm-1n(lnρa)
Therefore, the plot of ln(D) vs ln(m): slope=1n
Using the value of the slope and solving for n: n=1/slope
The error of ‘n’ can be solved by: Δn=nslope0+Δslope-nslope0
The power, ‘n’ is 3.3± 0.034 .. The expected value of 3 is near and within the error of the calculated power of the model equation.
The density of the material is defined as its mass per unit volume characterized often by the symbol rho (ρ). Different materials usually have different densities. In this lab, the densities of hand-made...
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