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Topics: Sphere, Volume, Radius Pages: 2 (306 words) Published: June 19, 2013
TrialVolume (CM3)Diameter (CM)Radius (CM)M&M Thickness (CM) 17511.345.67.743
28312.686.34.658

Table 2 – Direct Measurement
Trial M&M Thickness (CM) 1 .642
2 .741 3 .683

Table 3 - Calculated Averages
Method Calculated Average Thickness (cm) Indirect (from Table 1) .700 Direct (from Table 2) .689

1.When you performed Step 2 of the procedure, you actually made a cylinder of M&Ms. The cylinder was rather "smushed," and the height of the cylinder was the thickness of an M&M. Recall that the equation for the volume of a cylinder is V = (3.14)r2h. A. Rearrange the equation for "h." Show your work.

V = (3.14) r2h
V = (3.14) = 3.14r2h
(3.14)r23.14r2
V/(3.14)r2=h

B.Using the data from Table 1 and your equation, calculate the average thickness (height) of an M&M for each trial. Record your calculated values in Table 1. Hint: Students often forget that they must use the radius, and not the diameter, in the equation. Copy Table 1 into the assignment. C.Trial 1: h = 75 / (3.14 x 5.67)2

h = 75 / (3.14 x 32.1489)
h = 75 / 100.947546
h = 0.743 cm
Trial 2 h = 83 / (3.14 x 6.34)2
h = 83 / (3.14 x 40.1956)
h = 83 / 126.214184
h = 0.658 cm
C. You now have two values for the thickness of an M&M in Table 1. Determine the average M&M thickness using these values and record your value in Table 3. (0.743 + 0.658) / 2
= 1.401 / 2
= .700 cm
D. You have just determined a value for the thickness of an M&M using the indirect method. What makes this method "indirect"?
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