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279_Physics ProblemsTechnical Physics

# 279_Physics ProblemsTechnical Physics - Chapter 9 P9.69(a...

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Chapter 9 281 P9.69 (a) pFp if t += : 3 00 7 00 12 0 5 00 3 00 .. ± . ± kg m s N s kg bg b g ej af ji v f vi j f =+ 20 0 7 00 . ± . ± ms (b) a vv = fi t : a ijj i = +− = 20 0 7 00 7 00 500 400 . ± . ± . ± . . ± s 2 (c) a F = m : a i i == 12 0 . ± . ± N 3.00 kg 2 (d) rv a i tt 1 2 2 : rj i 700 1 2 2 . ± ± . s s 2 ri j 50 0 35 0 . ± . ± m (e) W =⋅ Fr : W + = 12 0 50 0 35 0 600 . ± . ± . ± N m m J ii j e j (f) 1 2 1 2 300 200 2 mv f + ± . ± . ± . ± kg m s 22 ij ij 1 2 150 449 674 2 mv f . k g J (g) 1 2 1 2 3 00 7 00 600 674 2 2 mv W i + = kg m s J J b g P9.70 We find the mass from Mt =− 360 2 50 kg kg s . . We find the acceleration from a M vdMd t MM M e = = Thrust kg s N 1500 250 3750 . We find the velocity and position according to Euler, from a t new old and xx v t new old If we take t = 0132 . s, a portion of the output looks like this: Time Total mass Acceleration Speed, v Position t (s) (kg) a 2 (m/s) x (m) 0.000 360.00 10.4167 0.0000 0.0000 0.132 359.67 10.4262 1.3750 0.1815
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