Problem 12B
Problem 12B
Problem 12B
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Givens7. h = 1.02 × 10 4 mL = 4.20 × 10 3 mk = 3.20 × 10 −2 N/mSolutionsF = kx = k(h − L)F = (3.20 × 10 −2 N/m)(6.0 × 10 3 m) = 190 N8. h = 348 mL = 2.00 × 10 2 mk = 25.0 N/mg = 9.81 m/s 2F = kx = k(h − L) = mgm =(25.0 N/m)(148 m)⎯⎯(9.81 m/s 2 )= 377 kgAdditional Practice <strong>12B</strong>1. L = 6.7 mg = 9.81 m/s 2 T = 2π ⎯ L g ⎯ = 2π ⎯ (9 ( .86.7m)⎯ 1 m / s2 =)5.2 sII2. L = 0.150 m(0.150 m)⎯⎯(9.81 m/s 2 )g = 9.81 m/s 2 T = 2p ⎯L g ⎯ ⎯ = 2p = 0.777 s3. x = 0.88 mg = 9.81 m/s 2T = 2p ⎯ 4 x⎯ =g4(0.88 m)2p ⎯⎯(9.81 m/s 2 )T = 3.8 s4. f = 6.4 × 10 −2 Hzg = 9.81 m/s 2 T = ⎯ 1 f ⎯⎯ = 2p ⎯L g ⎯ ⎯ g (9.81 m/s 2 )L = ⎯ 4p 2 ⎯f 2 = ⎯⎯⎯4π 2 (6.4 × 10 −2 Hz) 25. t = 3.6 × 10 3 sN = 48 oscillationsL = 61 mg = 9.81 m/s 2 T = 2p ⎯ L g ⎯ ⎯ = ⎯ N6. L = 1.00 mT = 10.5 st ⎯ ⎯ tN⎯2g(3.6 × 10 3 s) 2 (9.81 m/s 2 )L = ⎯ = ⎯⎯⎯4p 24p 2 (48) 2L = 1.4 × 10 3 mg = ⎯ 4p 2L 4p 2 (1.00 m)⎯T2= ⎯⎯ = 0.358 m/s 2(10.5 s) 2Copyright © by Holt, Rinehart and Winston. All rights reserved.II Ch. 12–2Holt Physics Solution Manual
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