s14 - Homework solution. Chapter 14. P14.6 (a) P = P0 + gh...

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Homework solution. Chapter 14. P14.6 (a) ( ) ( )( ) 53 2 0 1.013 10 Pa 1024 kg m 9.80 m s 1 000 m PP g h ρ =+ = × + 7 1.01 10 Pa P (b) The gauge pressure is the difference in pressure between the water outside and the air inside the submarine, which we suppose is at 1.00 atmosphere. 7 gauge 0 1.00 10 Pa PP P g h =− = = × The resultant inward force on the porthole is then () 2 75 1.00 10 Pa 0.150 m7 . 09 10 N FP A π ⎡⎤ == × = × ⎣⎦ . 2.00 m 1.00 m 2.00 m P14.14 The air outside and water inside both exert atmospheric pressure, so only the excess water pressure gh counts for the net force. Take a strip of hatch between depth h and h dh + . It feels force ( ) 2.00 m dF PdA . (a) The total force is 1.00 m h Fd F g h d = ∫∫ h FIG. P14.14 2 22 32 2.00 m1 0 0 0 k g m 9 . 80 m s . 00 m 29.4 kN to the right h Fg F = (b) The lever arm of dF is the distance ( ) h from hinge to strip: 33 . . 7.00 m3 . 1 000 kg m 16.3 kN m counterclockw ise h dg h h d h hh g ττ τρ τ = =− ⎢⎥ ⎛⎞ ⎜⎟ ⎝⎠ =⋅ 411
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412 Fluid Mechanics P14.17 0 Pg h ρ = () 5 0 33 2 10.13 10 Pa 10.5 m 0.984 10 kg m 9.80 m s P h g × == = × No. Som e alcohol and water will evaporate. The equilibrium vapor pressures of alcohol and water are higher than the vapor pressure of mercury. FIG. P14.17 P14.20 Let h be the height of the water column added to the right side of the U–tube. Then when equilibrium is reached, the situation is as shown in the sketch at right. Now consider two points, A and B shown in the sketch, at the level of the water–mercury interface. By Pascal’s Principle, the absolute pressure at B is the same as that at A . But, 0H g 2 01 2 and . Aw Bw PP g h g h PP g hh h ρρ =+ + ++ Thus, from , AB = 12 ww w w h h B A h h 1 h 2 water Mercury H g 2 h , or ( ) Hg 1 13. 611 .00 cm 12.6 cm w ⎡⎤ =− = = ⎢⎥ ⎣⎦ . FIG. P14.20 P14.25 (a) Before the metal is immersed: or 1 0 y FTM g =− = 2 1 1.00 kg 9.80 m s 9.80 N TM g = (b) After the metal is immersed: or 2 0 y FTBM g =+− = a scale b B M g T 1 M g T 2 2 w g BM g V g 3 1.00 kg 2 700 kg m M V Thus, FIG. P14.25 32 2 3 9.80 N1 0 0 0 k g m 9 .
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This note was uploaded on 08/31/2010 for the course PHYSICS 7E taught by Professor Staff during the Fall '08 term at UC Irvine.

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s14 - Homework solution. Chapter 14. P14.6 (a) P = P0 + gh...

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