173_Physics ProblemsTechnical Physics

173_Physics ProblemsTechnical Physics - Chapter 6 P6.45 (a)...

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Chapter 6 175 P6.45 (a) When vv T = , a = 0, Fm g C v T =− + = 2 0 v mg C T × × 480 10 980 250 10 13 7 4 5 .. . . kg m s kg m ms 2 ej e j (b) t s af x m v bg F mN a 2 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2 0 0 –0.392 –1.168 –2.30 –3.77 –5.51 –7.48 –9.65 –11.96 –14.4 0 –1.96 –3.88 –5.683 2 –7.306 8 –8.710 7 –9.880 3 –10.823 –11.563 –12.13 –12.56 – 4.704 – 4.608 – 4.327 6 –3.896 5 –3.369 3 –2.807 1 –2.263 5 –1.775 3 –1.361 6 –1.03 –0.762 –9.8 –9.599 9 –9.015 9 –8.117 8 –7.019 3 –5.848 1 –4.715 6 –3.698 6 –2.836 6 –2.14 –1.59 . . . listing results after each fifth step 3 4 5 –27.4 –41.0 –54.7 –13.49 –13.67 –13.71 –0.154 –0.029 1 –0.005 42 –0.321 –0.060 6 –0.011 3 The hailstone reaches 99% of v T after 3.3 s, 99.95% of v T after 5.0 s, 99.99% of v T after 6.0 s, 99.999% of v T after 7.4 s. P6.46 (a) At terminal velocity, g C v T == − + 0 2 C mg v T = × 22 4 0142 42 5 770 10 . . kg m s kg m 2 (b) Cv 24 2 360 0998 = . kg m m s N (c) Elapsed Time (s) Altitude
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