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What were the variables? . The independent variable was…. The dependent variable was…. The control variable was…. What were the variables?. The independent variable was….width of the polymer strip The dependent variable was….force needed to tear The control variable was….same plastic bag. - PowerPoint PPT Presentation
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What were the variables?
The independent variable was….The dependent variable was….The control variable was….
What were the variables?
The independent variable was….width of the polymer strip The dependent variable was….force needed to tearThe control variable was….same plastic bag
What is the interval of your independent variable is this suitable?
The interval was……..(with units).The interval was/was not suitable because….…again use your own results to justify your answer.
Mention: P (Yes/No), E (Pattern), D: (Data)
What is the interval of your independent variable is this suitable?
P: The interval was 0.5cm.
E: The interval was suitable because it showed me a clear pattern where as width increased, force required to tear the strip also increased. D: When the width was 0.5 cm … when the width was 1.5cm….Mention: P (Yes/No), E (Pattern), D: (Data)
Did you repeat any of your results?
P: Yes OR No?E: If yes then state why? Hint: Were your points close to the line of best fit? Did you have any anomalous results? Did you want to check for anomalous results? How was the mean calculated?D: Give examples from your results.If no, why not? Explain!
Did you repeat any of your results?
P: Yes E: I repeated my measurements to check for anomalous results which could be removed when I use my repeats to calculate a mean, reducing the effect of random error. D: For example, at 1.5 cm my first attempt was ____N whereas my second was ___N which fitted in with the pattern better.
Did you repeat any of your results?
P: No E: I didn’t repeat my measurements as there were no anomalous results which needed to be removed when calculating a mean, all of my results lied close to the line of best fit. D: For Example at 1.5 cm force required was ____N whereas my at 3.0 cm the force required was ____________ this pattern clearly supports my hypothesis.
Do your results support your hypothesis?P: Yes my results support the hypothesis. E: The greater the ……………….. the greater/ smaller the ……… needed to tear the polymer strip. D: 1.0cm needed a force of …………N.Whereas 3.0cm needed a force of ………..N.
Quote two sets of values from your results!
How do other people’s results compare to yours?
P: Are their results similar? E: State the pattern/ trend shown in their results in comparison to yours. Does it show the same trend? D: Give two examples from your and their results to justify your answer.
Sketch graph
From the data below draw a sketch graph…any comment on the interval used?
Width (cm) Force needed to tear (N)
1 71.5 9
2 113 154 18
Width of polymer strip (cm)
Force needed to tear strip (N)
Sketch graph
Does this data support your hypothesis?Case study 1
What do you notice about the data?
Width (cm)
Force needed to tear strip (N)
Try 1 Try 2 Try 3 Mean
1 24 26 28 26
2 41 44 47 44
3 66 62 64 64
4 88 111 89 96
Does this data support your hypothesis?Case study 1
Case Study 1 supports the hypothesis that as the width increases, so does the force needed to break the strip.
For example, the force for 1cm is 26N and the force for 4cm is 96N.
Does this data support your hypothesis?Case study 2
THICKNESS of polymer
strip (cm)
Force needed to tear strip (N)
Try 1 Try 2 Try 3 Mean
0.1 24 26 28
0.2 41 44 47
0.3 66 62 64
0.4 88 82 89
Does this data support your hypothesis?
THICKNESS of polymer
strip (cm)
Force needed to tear strip (N)
Try 1 Try 2 Try 3 Mean
0.1 24 26 28
0.2 41 44 47
0.3 66 62 64
0.4 88 82 89
Does this data support your hypothesis?Case study 2
Case Study 2 does not support the hypothesis because the independent variable is not the same. The changed the thickness whereas I changed the width of the polymer strip.
Does this data support your hypothesis?Case study 3
What do you notice about the data?
Width (cm)
Force needed to tear strip (N)
Try 1 Try 2 Try 3 Mean
1 24 26 28 26
2 41 44 47 44
3 66 26 64 52
4 88 111 89 96
Does this data support your hypothesis?Case study 3
What do you notice about the data?
Width (cm)
Force needed to tear strip (N)
Try 1 Try 2 Try 3 Mean
1 24 26 28 26
2 41 44 47 44
3 66 26 64 52
4 88 111 89 96
Does this data support your hypothesis?Case study 3
Case Study 3 supports the hypothesis. For example, the force for 1cm is 26N and the force for 4cm is 96N.However, there is an anomaly for the 3.0 cm width at Try 2 (26N). This has been used to calculate the mean.
Is it better to use a double layer handle instead of single layer handle?
Force (N)
Length of strip (mm)
Double layerSingle layer
10 30 50
10
20
30
How are the results from your experiment useful in the context given?
What is the context of this investigation ?
How could your results be used?
How are the results from your experiment useful in the context given?
What is the context of this investigation ?
A shopping company making bags could find the width of the handles needed to carry the force of the full bag.
How could your results be used?
A supermarket could find out force of a full carrier bag and use my results to calculate the minimum width the handles need to be to hold this force up.
