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0018 - 42 43 45 46 18(b At depth 10 meters the intensity...

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Unformatted text preview: 42. 43. 45. 46. 18 (b) At depth 10 meters the intensity is {(10) fined-4"") :(8.32x10”)10. (c) We solve the equation 1(x) = lee—"4‘ = 0.011", for x = (In 100)” .4 = 3.29 meters. -O_2J.r (a) The pressure at an altitude of x miles is given by p(x) = 29.923 . Hence the pressure at altitude 10000 ft is p(10000/5280) = 20.49 inches, and the pressure at altitude 30000 ft is p(300.0015280) = 9.60 inches. (1)) To find the altitude where p = 15 in., we solve the equation 29.92641” =15 for x = (1n 29.92/15)!0.2 m 3.452 miles = 18,200 ft. (a) A’ if rA-t-Q (b) The solution of the differential equation with 11(0) = O is given by Hi +Q = Qe”. When we substitute A = 40 (thousand), r = 0.11, and t = 18, we find that Q 2 0.70482, that is, $704.82 per year. 235 Let N80) and N50) be the numbers of 23st} and U atoms, respectively, at time t (in billions of years after the creation of the universe). Then N80) = Noe N50) = N057” , Where No is the initial number of atoms of each isotope. Also, 22 = (1n 2) I 4.51 and c = (1n 2)]0.71 from the given half—lives. We divide the equations for N8 and N5 and find that when I has the value corresponding to "now", 7’“ and etc-*t' = £3 2 137.7. N5 Finally we solve this last equation for r = (ln137.7)/(cr— k) z 5.99. Thus we get an estimate of about 6 billion years for the age of the universe. The cake's temperature will be 100° after 66 min 40 see; this problem isjust like Example 6 in the text. (b) By separating the variables we solve the differential equation for c— rP(r) = (c— rPc.) 6”. With P0) = 0 this yields Chapter 1 ...
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