In this investigation, the mass of chewing gum after chewed at different times were recorded by using an electronic balance. The average final mass was calculated and then graphed with the time. The graph tells us that an inversely proportional relationship exists between the final mass and the time. The difference between final and initial mass was also calculated and was graphed with time, which gave us a linear graph. Through this graph it is found that the longer you chew the less mass it gets.

This is because sugar, the main ingredient in the gum, gets lost from the gum at a rate 0.0108 + 0.0005 g/s. This result is found from the slope of a liner graph which in this case tells the proportionality factor between avg. difference of mass and the time.Evaluation: The lab was a success. This can be seen form the linear relationship between average difference of mass of the gum and the rate of decomposition. But there were some minor problems that occurred during the lab, as it can be seen that the slope varies about + 0.

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0005. One problem could have been that the gum when weighed after it was chewed was sticking to its wrapper. This caused it to gain more mass because the tiny bits of wrapper might got stuck in the gum thus increasing its mass.Improvements: The problem could be solved by using the weigh boat instead of the wrapper. This could have made the gum not stick in the surface.

In this experiment I investigated the propagation of errors while calculating the volume of two objects. I came to the conclusion that a measuring instrument like a screw gauge has some problematic limitations. This, along with other human factors, introduces uncertainties and errors in the measurement. Careful procedure can minimize these errors but cannot remove them completely. The errors in individual measurements contribute to the result of calculations using the measured quantities.

Various precautions need to be taken to minimize errors in measurements and study of how these errors propagate during the various calculations needs to be taken.The aim of this experiment was to investigate the propagation of errors while calculating the volume of a Cylindrical and a Spherical object. My experiment included calculating the volume of two objects by two different methods. Apparatus: Vernier calipers, given spherical object, given cylindrical object, a measuring cylinder, and string. In the beginning of this experiment I calculated the volume of both the given objects by two different methods each. I decided to first take the spherical object.The amount of water displaced by the spherical object is equal to the volume of said object.

The rise in the level of water came out to be 9cm3. The error in this case can be calculated as + 1 (because the LC of the cylinder is 0.5 mm). Next we calculate the volume using the formula to calculate the volume of a sphere i.e.

Volume of a sphere = 4/3 r3. The radius of the spherical object is half the diameter, which I found using the Vernier Calipers. The final volume is 9.

525 cm3. The error of the volume is thrice the percentage error of the radius/diameter. This is because the radius is cubed; hence the error has to be multiplied by 3. The final value of the error of the spherical object is +0.108 cm.

Thus V1 = 10 cm3 + 1 cm and V2= 9.525cm3 + 0.108cm.We use similar methods for calculating the volume of the cylindrical object.

First, by calculating the amount of water held in the object (with the help of a measuring cylinder). The volume = 310 cm3 + 5cm3(as the least count of the cylinder is 2.5 cm). Now we calculate the volume by using the formula for volume of a cylinderVolume = r2h. The final value I got was305.

614 cm3. The error in this case is twice the percentage error of the radius (as the radius was squared). This gives us a resultant error of + 0.096 cm3.

Thus we get the final volume of the cylinder to be =305.61 cm3 + 0.096 cm3.I noticed that there is a big difference between both the methods (with the sphere and the cylinder). E.

g.: – volume of sphere with water displacement is 10 +1 cm2, but when we use the formula it comes out to be 9.525 + 0.108 cm2. This shows us that with the correct precautions and methods, we can get an answer that has the least possible error and is therefore the most accurate. My expectation regarding the difference was hence right.

I found this to be an engaging and eye-opening experiment and I enjoyed doing it.

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