{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}