Create a processed data table if more than one trial has been performed. Include formulas and sample calculations below the data table. 3. Make a graph to show the relationship between the independent and dependent variables.

Figure I: Average number Of paperclips per set Of loops Follow-up Questions: 1. How do you think the results would have been affected if the iron nail used had zero loops of copper wire wrapped around it? Justify. It wouldn’t have worked since It wouldn’t have created a magnet because you wouldn’t have a core wrapped with wire and it wont be magnetized.How do you think the results would have been affected if the nail had been made of aluminum? Just;’. Aluminum is not a ferromagnetic metal therefore it wouldn’t have worked.

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3. Using the graph, estimate the number of paperclips that the electromagnet would have picked up if it had 30 loops of copper wire wrapped around it. Show your work (extrapolation using the best fit line to estimate your answer) on the graph. Every set of loops it increases by 2. 5 paperclips, therefore if you have 30 loops it would be 8 paperclips. Conclusion:Repeat steps 4-7 for a total of S trials, 9.

Repeat steps 3-8 by increasing the potential difference by IV each time. Table 3: Measuring the amount of paperclips when the voltage of the power pack is raised by 1 every trial. Voltage (V) Current Intensity (A) Trial 5 3. 9 14 10 11 Table 3: Averaging the numbers of paperclips per voltage (25 loops) Current Intensity (A) 2. 2 110 2. Average (trial *trial trial 34 trial 44 trial 5 = answer / # of trials = average) 11/5 2. 2 Refer to Part A. Figure 2: Average number Of paperclips per rate Of current intensity Follow-Up Questions: 1.

Using the graph, estimate the number of paperclips that the electromagnet would have picked up if the current intensity of the circuit was 5. 0 A. Show your work (extrapolation) on the graph. If the current intensity was 5. 0 A It would be 8 paperclips because it continues the curve. 2.

An electromagnet can pick up six paper-clips. Fifth current is halved and the number of loops is tripled, how many paperclips do you expect the electromagnet to lilt? Show your work. 6 PC. Current is halved: 3 PC.

Loops multiplied by 3: 9 PC. Increase its strength. (it will also pick up more paperclips)