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Pennies And Dimes Lab Report

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Discussion
As presented in Table 1, only pennies were used in this part of the experiment. 8 pennies were added to the cup at each interval, and in response to the addition, the force consistently increased by 0.20 N. This demonstrated an upward trending linear relationship, as can be seen in Figure 1. The slope, or m, represents the weight of a penny, which, is shown in Figure 1 as 0.025 N. The x-values represent the independent variable, which is the # of pennies, while the y-values represent the dependent variable, force (N). Slope can be represented as the change in the force (N) divided by the change in the # of coins. The value for slope in this data set represents the force per coin. The y-intercept value (b) is 0, which is the y value …show more content…

Initially, 8 pennies were added to the cup, followed by the addition of 7 pennies and 1 dime, then 4 pennies and 4 dimes, and finally 8 more pennies. There were therefore a total of 27 pennies and 5 dimes added to the cup. Table 2 demonstrates that the force (N) for dimes and pennies went up by almost 0.20 N at each interval. Therefore, the force (N) in Table 2 did not deviate much from the force (N) seen in Table 1 where all pennies were used. The reason little variation in force was seen in Table 2 was due to mostly pennies being added to the cup. Due to so many pennies being added, the dimes had little impact on the overall force (N). If roughly an equal ratio of pennies to dimes had been added to the cup, a more distinct variation between the slope’s in Figure 1 and Figure 2 would have been seen. However, the slope or the average weight of the coins, as represented in Figure 2, was 0.0249 N. The slope can be calculated by dividing the change in the force by the change in the # of pennies and dimes. The x-values represent the number of pennies and dimes, while the y-values represent force (N). The y-intercept value is equal to 0, therefore, the linear equation is y=0.0249x+0. After plotting the line on a manual curve fit, as can be seen in Figure 2, the R2 value was 0.99884. This R2 value is very close to 1, meaning that the match of the linear model to the data fits. After running a …show more content…

Initially, 1 dime was added to the cup, followed by the addition of 1 penny, then 1 nickel, and finally 1 quarter. The idea was to add the coins in order from the lightest coin in weight (dime) to the heaviest (quarter) in order to get an upward trend line. When 1 dime was added to the cup, the force was 0.03 N, however, when a penny was added to the cup, the force was increased to a total of 0.05 N. This represented only a 0.02 N change when the penny was added. This information is inferring that a dime weighs more than a penny, however, this is not true. An error that may have occurred could be that the penny was added before the dime was added. Therefore, the penny would have weighed 0.3 N and the dime would have weighed 0.2 N. However, the information in Table 3 does demonstrate that the nickel weighs less than the quarter, but more than the penny and dime, giving an upward trend line. The slope of the line passing through points (1,0.03) and (4,0.2) is 0.056 N. Therefore, 0.056 N is the average weight of the coins. The x-values represent the # of coins, the y-values represent the force (N), and the y-intercept value is 0. Initially, a linear graph was generated for Figure 3, however, the trend line did not pass through each point on the graph and the R2 value was .82. Therefore, a polynomial trend line was incorporated into the graph, as seen in Figure 3. The

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