Answer Key
835
ds =4 and ∫ 2 4
213 . P ( s ) =4 s , so dP
4 ds =8.
2
215 . ∫
1 Nds = N 217 .With p as in the previous exercise, each of the 12 pentagons increases in area from 2 p to 4 p units so the net increase in the area of the dodecahedron is 36 p units. 219 . 18 s 2 =6 ∫ s 2 s 2 xdx 221 . 12 πR 2 =8 π ∫ R 2 R rdr 223 . d ( t ) = ∫ 0 t v ( s ) ds =4 t − t 2 . The total distance is d (2) =4m. 225 . d ( t ) = ∫ 0 t v ( s ) ds . For t <3, d ( t ) = ∫ 0 t (6−2 t ) dt =6 t − t 2 . For t >3, d ( t ) = d (3)+ ∫ 3 t (2 t −6) dt =9+( t 2 −6 t ) | 3 6 . The total distance is d (6) =18m. 227 . v ( t ) =40−9.8 t m/sec; h ( t ) =1.5+40 t −4.9 t 2 m/s 229 . The net increase is 1 unit. 231 . At t =5, the height of water is x = ⎛ ⎝ 15 π ⎞ ⎠ 1/3 m.. The net change in height from t =5 to t =10 is ⎛ ⎝ 30 π ⎞ ⎠ 1/3 − ⎛ ⎝ 15 π ⎞ ⎠ 1/3
m. 233 . The total daily power consumption is estimated as the sum of the hourly power rates, or 911 gW-h. 235 . 17 kJ 237 . a. 54.3%; b. 27.00%; c. The curve in the following plot is 2.35( t +3) e −0.15( t +3) .
239 . In dry conditions, with initial velocity v 0 =30 m/s, D =64.3 and, if v 0 =25, D =44.64. In wet conditions, if v 0 =30, and D =180 and if v 0 =25, D =125. 241 . 225 cal 243 . E (150) =28, E (300) =22, E (450) =16 245 . a.
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