Relative uncertainty0.032258064 As we are finding the volume, we need to add the uncertainties of all the measurements, i.e. length, width and height. In this case the material was a cube, where the measurements were the same and even the uncertainties, thus we can multiplied the uncertainty by 3, i.e. 0.

032258064 x 3 = 0.0967741935 When converted in percentage we get: 9.677% 10% Volume = 3.

1cm x 3.1cm x 3.1cm = 29.791 cm3 Converting it to two significant figures, we obtain, Now add the uncertainties, i.e. 0.

0004 + 0.1, which gives 0.1004, which if rounded off to one significant figure is 0.1. Dividing 251.7 by 30, we get 8.39.

Thus, (8.39 ± 0.1) g cm-3 = (8.4 ± 0.

1) g cm-3 Below are the results for the readings for the aluminum materials: The calculations for finding the volume and density are done in the same way as the before sample calculation shows. The above graph shows the density graphically. If we notice then the slope of the black line is all most equal to the average of the three densities of the aluminum materials.

Thus, the average density is equal to the gradient of the black line in the graph. The other two lines show the minimum (red line) slope of the density and the maximum (green line). It is not necessary that their average will be the slope of the black line. Aim: To investigate the bending of a cantilever beam and to determine the mathematical constants involved in the equation x = kln Hypothesis and Theory:The depression of a cantilever beam is given by the equation The depression of a cantilever beam, is related to its length,, by the equation , where The length of the beam varies with the depression according to the mass that has been used. If we plot a graph with these variables we will get a curve, but it is not clear from this curve what the values of k and n work out to be. On top of this, if we do not know what the value of n is.So, we take logs of both sides of the equation.

The equation below has used natural logarithms: ln () = ln ( ln () = ln () + ln () ln () = ln () + ln () This is now in the same form as the equation for a straight-line: Thus, if we plot ln () on the y-axis and ln () on the x-axis we will get a straight-line graph. The gradient will be equal to. The y-intercept will be equal to ln () [so =] Since the log graph is going to be a straight line graph in that case it will be easy to find the mathematical constants which are the aim of the experiment. Apparatus: Two meter rules, adjustable clamp, weights, and a metre ruler.

x

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