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Unformatted text preview: Nina Fitch SID#17849531 1) A) Vaisala frequency: N 2 = (g / )(d / dz) = sigma T + 1000 d /dz = d(sigma t)/dz d /dz max = d(sigma t)/dz max, where the slope of sigma t is the least negative sigma t is the least negative between the values of 125 meters and 175 meters where the slope 45 from the graph = sigma t + 1000 27 Z dp/dz 50 720 75 720 100 720 125 90 150 45 175 90 200 880 225 880 250 880 275 880 300 880 325 880 350 880 375 880 400 880 425 880 450 880 475 880 500 880 525 880 550 880 575 880 600 880 N 2 = (g / )(d / dz) = (9.81m/s / 27) x 45 4 B) The buoyancy period, T N = 2 /N= 2 /4 = 1.6 2) I = 9 N = 20 A= I /3 = 3 B= N /10 = 2 P = 3 Z I(z) N(z) Pg 0 9 0 0 1 8.143536762 0.65567799 0.54128558 2 7.368576778 1.289860299 0.835893483 3 6.667363986 1.903251639 1.008874188 4 6.032880414 2.496533619 1.112446886 5 5.458775937 3.070365502 1.172356816 6 4.939304725 3.625384938 1.202836183 7 4.469267734 4.162208673 1.212458964 1....
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This homework help was uploaded on 04/04/2008 for the course EPS 106a taught by Professor Bishop during the Spring '08 term at University of California, Berkeley.
 Spring '08
 Bishop

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