( )
5
pool
32
27
(10/18) 4.32 10 kg
(10 / 18)
1.44 10 .
1.67 10
kg
p
M
N
m
−
×
==
=
×
×
Using Eq. 4220, we obtain
R
N
T
×
≈
ln
.l
n
.
/
2
144 10
2
10
1
12
32
32
ch
y
decay y
4. Since the density of water is
ρ
= 1000 kg/m
3
= 1 kg/L, then the total mass of the pool is
V
= 4.32
×
10
5
kg, where
V
is the given volume. Now, the fraction of that mass made up
by the protons is 10/18 (by counting the protons versus total nucleons in a water
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This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Mass

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