11_Physics ProblemsTechnical Physics

11_Physics ProblemsTechnical Physics - Chapter 1 P1.40 11...

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Chapter 1 11 P1.40 The mass of each sphere is m V r Al Al Al Al Al = = ρ π ρ 4 3 3 and m V r Fe Fe Fe Fe Fe = = ρ π ρ 4 3 3 . Setting these masses equal, 4 3 4 3 3 3 π ρ π ρ Al Al Fe Fe r r = and r r Al Fe Fe Al = ρ ρ 3 . Section 1.6 Estimates and Order-of-Magnitude Calculations P1.41 Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ball as a sphere of diameter 0.038 m. The volume of the room is 4 4 3 48 × × = m 3 , while the volume of one ball is 4 3 0 038 2 87 10 3 5 π . . m 2 m 3 F H G I K J = × . Therefore, one can fit about 48 2 87 10 10 5 6 . ~ × 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 the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called “best packing fraction” is 1 6 2 0 74 π = . so that at least 26% of the space will be empty. Therefore, the
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