P18_068 - for K , we obtain I = 1 . 5 W/m 2 . (c) The...

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68. (a) Using m =7 . 3 × 10 7 kg, the initial gravitational potential energy is U = mgy =3 . 9 × 10 11 J, where h = 550 m. Assuming this converts primarily into kinetic energy during the fall, then K = 3 . 9 × 10 11 J just before impact with the ground. Using instead the mass estimate m =1 . 7 × 10 8 kg, we arrive at K =9 . 2 × 10 11 J. (b) The process of converting this kinetic energy into other forms of energy (during the impact with the ground) is assumed to take ∆ t =0 . 50 s (and in the average sense, we take the “power” P to be wave-energy / t ). With 20% of the energy going into creating a seismic wave, the intensity of the body wave is estimated to be I = P A hemisphere = (0 . 20) K/ t 1 2 (4 πr 2 ) =0 . 63 W / m 2 using r = 200 × 10 3 m and the smaller value for K from part (a). Using instead the larger estimate
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Unformatted text preview: for K , we obtain I = 1 . 5 W/m 2 . (c) The surface area of a cylinder of height d is 2 rd , so the intensity of the surface wave is I = P A cylinder = (0 . 20) K/ t (2 rd ) = 25 10 3 W / m 2 using d = 5 . 0 m, r = 200 10 3 m and the smaller value for K from part (a). Using instead the larger estimate for K , we obtain I = 58 kW/m 2 . (d) Although several factors are involved in determining which seismic waves are most likely to be detected, we observe that on the basis of the above Fndings we should expect the more intense waves (the surface waves) to be more readily detected....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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