Chapter 1
11
P1.40
The mass of each sphere is
mV
r
Al
Al
Al
Al Al
==
ρ
πρ
4
3
3
and
r
Fe
Fe Fe
Fe Fe
4
3
3
.
Setting these masses equal,
4
3
4
3
33
Al Al
Fe Fe
rr
=
and
Al
Fe
Fe
Al
=
3
.
Section 1.6
Estimates and OrderofMagnitude Calculations
P1.41
Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each pingpong ball
as a sphere of diameter 0.038 m. The volume of the room is
4434
8
××=
m
3
, while the volume of
one ball is
4
3
0038
287 10
3
5
π
.
.
m
2
m
3
F
H
G
I
K
J
=×
−
.
Therefore, one can fit about
48
10
5
6
.
~
×
−
pingpong 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 socalled “best
packing fraction” is
1
6
20
7
4
=
.
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 .
 Fall '11
 Staff
 Physics, Mass

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