CH14 - 10 9 N 4. dF = dm 2 r = (2 r h dr) 2 r = dP(2 rh),...

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2. V = (100 g)(1 g/cm 3 ) =100 cm 3 . Mass of liquid = 115 g, thus ρ = 115/100 = 1.15 g/cm 3 . 5. A = (Y/F)(ΔL/L) = π D 2 /4 leads to D = 1.75 mm 6. B = –ΔP. V/ΔV = +1.05 × 10 9 N/m 2 12. (a) mg = PA, so m/A = P/g = 10.3 × 10 3 kg/m 2 . Since A = 4 π R 2 , M = 5.25 × 10 19 kg. (b) V = 4 π /3 [(R + h) 3 – R 3 ] = 4 π R 2 h m = ρ V, thus h = 8 km. 30. W = (600)(3 ×  3 ×  0.16)g = 864g = 9.47 × 10 3 N F B = ρ f gV = (10 3 )(9.8)(0.8 × 9 ×  0.16) = 11.3 × 10 3 N Thus load is 2.82 × 10 3 N or 288 kg. 36. mg = (950)g(0.12h), thus h = 12.6 cm 41. W = 3(1025)A = 1000Ah, h = 3.075, so Δh = 7.5 cm. 47. A 1 V 1 = A 2 v 2 , v 2 = (2.4)(1.59/1.28) 2 = 3.7 m/s 51. (a) A 1 v 1 = A 2 v 2 , so v 2 = 3.6 m/s (b) P 1 + 1/2 ρ v 1 2 = P 2 + ρ gh + 1/2 ρ v 2 2 Find P 1 = 370 kPa. 53. v 2 = (1.59/1.28) 2 (2.4) = 3.7 m/s. P 1 + 1/2 ρ v 1 2 = 1/2 ρ v 2 2 , so P 1 = 3.79 kPa. Problems 1. dF = W dy( ρ g) (H – y); F = ρ gW (Hy – y 2 ) dy = ρ gWH 2 /2 F = 3.5
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Unformatted text preview: 10 9 N 4. dF = dm 2 r = (2 r h dr) 2 r = dP(2 rh), thus dP = 2 r dr. After integration : P = P a + 2 r 2 /2 5. m = s (2 Rt) (L + r), and F B = mg = f ( R 2 h)g Fraction = h/L = 2 s t(R + L) / RL f 8. A 1 v 1 = A 2 v 2 gives R 2 v o = r 2 v v 2 = v o 2 + 2gy thus (R 2 v o 2 /r 2 ) 2 = v o 2 + 2gy r 4 = R 4 [1 + 2gy/v o 2 ] 1 10. P B + 1/2 f v B 2 + f g = p A + f gh, so P = 1/2 f v B 2 Also, P = m gh, thus v B = (2 m gh/ f ) 1/2 11. P 1 + 1/2 v 2 + gy + P 2 + 1/2 v 2 ; P 1 = P 2 + P o . v 2 = 2gy = 2g (1 h) y =1/2 gt 2 , thus t = 0.09 s; R = vt; t = (2H/g) 1/2 , thus R = 2 [h(H h) 1/2 ]...
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CH14 - 10 9 N 4. dF = dm 2 r = (2 r h dr) 2 r = dP(2 rh),...

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