ch44-p004 - 4 Since the density of water is = 1000 kg/m3 =...

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( ) 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. 42-20, 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.

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