HW 5 solutions (Ch 15)

# HW 5 solutions (Ch 15) - Physics 6C HW#5 solution May 8...

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Physics 6C HW#5 solution May 8, 2008 15.6 a) P = P 0 + ρgh = 1 . 013 × 10 5 Pa + (1024 Kg/m 3 )(9 . 80 m/s 2 )(1000 m ) = 1 . 01 × 10 7 Pa b) The gauge pressure is the diﬀerence in pressure be- tween the water outside and the air inside the sub- marine, which we suppose is at 1.00 atm. P gauge = P - P 0 = ρgh = 1 . 00 × 10 7 Pa The resultant inward force on the porthole is then. F = P gauge A = 1 . 00 × 10 7 Pa ( π × (0 . 150 m ) 2 ) = 7 . 09 × 10 5 N 15.8 a) Suppose the “vacuum cleaner” functions as a high- vacuum pump. The air below the brick will exert on it lifting force. F = PA = 1 . 013 × 10 5 Pa ( π (1 . 43 × 10 - 2 m ) 2 ) = 65.1 N b) The octopus can pull the bottom away from the top shell with a force that could be no larger than F = PA = ( P 0 + ρgh ) A = ± 1 . 013 × 10 5 Pa + (1030 Kg/m 3 )(9 . 80 m/s 2 )(32 . 3 m ) ² × ( π (1 . 43 × 10 - 2 m ) 2 ) = 275 N 15.14 a) Using the deﬁnition of density, we have h w = m water A 2 ρ water = 100 g 5 . 00 cm 2 (1 . 00 g/cm 3

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HW 5 solutions (Ch 15) - Physics 6C HW#5 solution May 8...

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