Chapter 111P1.40The mass of each sphere ismVrAlAlAlAl Al==ρπρ433andrFeFe FeFe Fe433.Setting these masses equal,434333Al AlFe Ferr=and AlFeFeAl=3.Section 1.6Estimates and Order-of-Magnitude CalculationsP1.41Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ballas a sphere of diameter 0.038 m. The volume of the room is 44348××=m3, while the volume ofone ball is430038287 1035π..m2m3FHGIKJ=×−.Therefore, one can fit about 481056.~×−ping-pong balls in the room.As an aside, the actual number is smaller than this because there will be a lot of space in theroom that cannot be covered by balls. In fact, even in the best arrangement, the so-called “bestpacking fraction” is 162074=.so that at least 26% of the space will be empty. Therefore, the
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .