calculations - 1. Diameter of Sphere and Computation of its...

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1. Diameter of Sphere and Computation of its Area and Volume The area of a sphere of radius, r , is given by A = 4 π r 2 . The volume, by (4/3) π r 3 . Since the diameter, d , is twice the radius, ( d = 2 r )we can rewrite the above as: A = π d 2 and V = (1/6) π d 3 Suppose our sphere is a basketball and we measure its diameter with a ruler whose smallest division is 1 inch. (Such a ruler is shown below.) Using our rule of thumb, we can read the ruler to the nearest 0.2 in.). We measure the diameter to be 9.4 in. Our understanding is that our diameter may have an error of ±0.2 in. due to our visual estimation which corresponds to a percent error of 100 X 0.2 / 9.4 = 2.1%. We seek to find out what the impact of such an error will be on the area and volume of the ball. A simple way is to calculate the area and volume of the sphere from the value of our measured diameter, and then to add and subtract the uncertainty and recalculate the area and volume. Diameter (in) Area (in 2 ) Volume (in 3 ) We write the Area and Volume in scientific notation because of the number of signficant figures justified in their values 9.4 2.8 X 10 2 4.4 X 10 2 9.4 - 0.2 = 9.2 2.7 X 10 2 4.1X 10 2 9.4 + 0.2 = 9.6 2.9 X 10 2 4.6 X 10 2 The Range of the volume (the difference between the maximum and minimum values) is 4.6 X 10 2 - 4.1 X 10 2 = 5 X 10 1 (I.e., ~50 in 3 separate the minimum and maximum values!). Note that
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calculations - 1. Diameter of Sphere and Computation of its...

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