ex2soln_w07

# ex2soln_w07 - Physics 106 Exam 2 Solutions 1 Convert the ow...

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Physics 106 - Exam 2 Solutions 1. Convert the flow rate to cubic meters per second, then use power equals flow rate times g times height: 3300 ft 3 s × ( 1 m 3 . 28 ft ) 3 = 93 . 52 m 3 s [93 . 52 m 3 s × 10 3 kg m 3 ] × 9 . 81 m s 2 × 10 m = 9 . 17 × 10 6 J s = 9 . 17 MW 2. r = d 2 = 17 . 5 m The total power in the wind is: π × (17 . 5) 2 × (6 . 1 × 10 - 4 ) × (20 3 ) = 4 , 700 kW = 4 . 70 MW The total electrical power, accounting for the efficiencies of conversion of wind energy to rotor KE and conversion of rotor KE to electrical power, is: (4 . 70 MW ) × 0 . 59 × 0 . 75 = 2 . 08 MW 3. The total input power into the OTEC device is: (2 × 10 3 kg s ) × ( 10 3 g kg ) × 2 . 5 C × 1 cal g C × 4 . 184 J cal = 20 . 92 MW The efficiency of the OTEC heat engine, using the surface and subsurface water temperatures in K, is: η = 1 - 277 299 = 0 . 0736 Multiplying these two factors, and including the electrical conversion efficiency, gives the power output: 20 . 92 MW × 0 . 0736 × 0 . 9 = 1 . 39 MW 4. 174 71 Lu 103 174 70 Y b 104 + β + + ν 5. 16 × 10 9 250 × 10 6 = 64 = 2 6 Thus, there are 6 half lives, so T = 6 × 3 . 31 yr = 19 . 9 yr 6. Neutrons (a) are absorbed by

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