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Nt1310 Unit 3 Configuration

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Interpretation
In order to interpret the data, first it was categorized based on the number the ball had been assigned in its configuration, the number of holes and the size of holes. There were ten configurations with three balls per configuration. Each ball in the configuration was assigned a number. Each ball had been dropped ten times between the photogates for a total of 300 drops. Each drop was charted in our data.
The results for the ping pong balls with zero holes were analyzed first. Next, the analysis of the other three sets of hole configurations, two, four and six holes, was completed. After this was finished, the three sets of hole sizes were analyzed. This included the ping pong balls with 3/16”, ¼”, and 5/16” holes. …show more content…
Using the Anova tables the data was organized by the number of drops of each ping pong ball as well as by the hole size. The sum of the time of free fall for each drop was used and the average taken. The values for each ping pong ball was determined and solved. When comparing the balls with the 3/16”, ¼”, and 5/16” holes, there was not a significant difference. However, the ball that had 5/16” holes had the longest free fall time of 0.3570 seconds. When comparing this value to the values of the other ball configurations in our experiment, this was the ball with the most drag or air resistance.To compare how the size of each hole affected the time a ball took to drop, a means plot was created. (Figure x) It can be seen from Figure 2 that when comparing balls with the same number of holes but with different sizes of holes, there is a slight change in the time it took the balls to drop between the photogates. In this case, when comparing the largest and smallest holes there is a slight decrease in time. This indicates that the larger the holes, the greater the air

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