Compare the slope of the
velocity-time graph to the average of all your acceleration values. Are they
close? What does the slope of a velocity (or speed) vs. time graph mean?
Explain the answer using your data. How does the value of g that you calculated compare to the accepted value of 9.80 m/s2?
What is your percent error? Remember that the value of g can be calculated by using g
= a/sinθ and percent error can be calculated using the
following equation the equation is % error= ((measured value-accepted value)/accepted value) x 100 I tried solving this but only got this:
% error= (0.1846 – 9.8)/9.8) x 100I uploaded the data table from my experiment and i hope somone would lend me some help because ‘;m stuck!1.25
Tl T2 T3
Aug. Augof Vebcity
IM
1.33 1.15s 1.25 0.8m/s
6.9M 0.95 lills 1.33 1.ls 0.8mls
0.8M 0.93 0.985 1.01 0.99 0.8mls
0.71 0.965 0.815 0.845 35 0.2wls
061 10.8ls 0.755
0.83 35
0.2m/s
0.5M 0.65 0.665 0.65 0.6 0.8wls
0.4M/0.60, 0.65 0.653, o.lols
0.39 /0.63,-0.60 0.40s 13s 0.2m6
0.2M 0.40 0.368 0.38 1.145
o.lmls
OIM 10.403|0.33 0.28 (0.3, 03ml
Aug of Acc.
0.6m/s
0.7ml
0.9m13
097 00mis
0.06m/s
1.3m/s
0.04m/s²
0.1mls
0.15ml5
lillnls
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