Sol-Ch14 - CHAPTER 14 E xercises 1. (a) m = (0.7v)(0.8) +...

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Exercises 1. (a) m = (0.7v)(0.8) + (0.3V)(1) = 0.86V, thus ρ = 0.86 g/cm 3 (b) V = (0.7m)/0.8 + (0.3m)/1. Then ρ = m/V = 0.851 g/cm 3 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 . 3. (a) ρ = m/v where v = 4πr 3 /3. ρ = 8.95 × 10 14 kg/m3 (b) v = 4πr 3 /3 = M/ ρ , find r = 1.17 km 4. Y = FL/(A ΔL) = 6.677 × 10 10 N/m 2 5. A = (Y/F)(ΔL/L) = πD 2 /4 leads to D = 1.75 mm 6. B = –ΔP V/ΔV = +1.05 × 10 9 N/m 2 7. A = (Y/F)(ΔL/L). F/A = 1.2 × 10 8 N/m 2 where F = m(g +a) = 9040 N. then A = 9040/1.2 × 10 8 = πD 2 /4 leads to D = 9.8 mm. 8. ΔV/V = –ΔP/B = –5.14 × 10 –2 9. F = AYΔL/L, thus F = (3 × 10 –4 )(10 10 )(10 –2 ) = 3 ×  10 4 N 10. F = (πd 2 /4)S = 3.96 × 10 4 N 11. F = (πdh)S = 1.06 × 10 4 N 12. (a) mg = PA, so m/A = P/g = 10.3 × 10 3 kg/m 2 . Since A = 4πR 2 , M = 5.25 × 10 19 kg. (b) V = 4π/3 [(R + h) 3 – R 3 ] = 4πR 2 h m = ρ V, thus h = 8 km. 13. F = ΔP A = (0.6 × 101.3 kPa)(1.68 m 2 ) = 1.02 × 10 5 N 14. F =ΔP A = 600 N 15. F = PA = (20/760)(1.013 × 10 5 N/m 2 )(π)(5 ×  10 –3 m) 2 = 0.21 N 16. P= P o + ρ gh = 1.013 × 10 5 + 9800h CHAPTER 14
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(a) 131 kPa ; (b) 1.08 MPa; (c) 107 MPa 17. Let y = 0 at the mercury-water interface in one arm. The pressure is the same at this level in the other arm containing just Hg. V
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This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.

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Sol-Ch14 - CHAPTER 14 E xercises 1. (a) m = (0.7v)(0.8) +...

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