Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 1 - Measurement - Problems - Page 10: 24


0.26 kg

Work Step by Step

The general idea of what we want to do here, is find out how many grains of sand (spheres) it would take to equal the surface area of the cube described, calculate the volume of all those spheres, then find the mass from that volume with the given density. First, as stated above, we need to find how many spheres it would take to equal the surface area of the cube. We can do this by simply taking the ratio of the surface area of the cube, to the surface area of one of the spheres. We know a cube has equal side lengths, and the cube described has a side with a length of 1 meter. So lets find the surface area. You can use the formula of the surface area for a cube which is SA = 6A 6 sides * 1 meter * 1 meter = 6 $m^{2}$ SA = 6 $m^{2}$ Now lets figure out the surface area of the grain of sand (sphere) with a radius of 50 µm. The formula for the surface area of a sphere is SA = 4πr$r^{2}$ We can use the conversion of 1 µm = 1.0 * 10$^{-6}$ m to convert the radius to meters r = 50 µm = 50 * 10$^{-6}$ m SA = 4π(50 * 10$^{-6}$ m)$^{2}$ SA = 3.1416 * 10$^{-8}$ $m^{2}$ In order to find out how many spheres would it take to equal the surface area of the cube, we simply have to divide the surface area of the cube, by the surface area of the sphere. Number of spheres = $\frac{Surface Area of the Cube}{Surface Area of the sphere}$ Number of spheres = $\frac{(6) m^{2}}{(3.1416 * 10^-8) m^2}$ Number of spheres = 190,985,485.103 or 1.91*10 $^{8}$ Now that we know how many spheres we are dealing with, we can strictly work with the density formula which is p = $\frac{m}{V}$, where p is density. You can rearrange this formula so we can find the mass of the spheres, which will give you m= pV. The density (p) of the spheres was given to us as 2600 $\frac{kg}{m^3}$, but we don't know the volume. So lets calculate that, then we will be able to find the mass needed. Volume of a Sphere V = $\frac{4}{3}$ πr$^{3}$ V = $\frac{4}{3}$ π(50*10$^{-6}$)$^{3}$ V = (5.2360 * 10$^{-13}$) m$^{3}$ This is the volume for only one grain of sand. We now have to multiply this number by all the grains of sand to find their total volume. Total Volume of Grains of Sand Volume of a Grain of Sand * Number of Grains (5.2360 * 10$^{-13}$) m$^{3}$ * (1.91*10 $^{8}$) grains Total Volume = (1.0 * 10 $^{-5}$) m$^{3}$ Now we can finally find the mass of all those sphere using dimensional analysis or m = pV Mass of Grains of Sand m = pV m = 2600 $\frac{kg}{m^3}$ * (1.0 * 10 $^{-5}$) m$^{3}$ m = 0.26 kg
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