P15_059

# P15_059 - f = 0 90 we Fnd the apparent weight is zero(it...

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59. (a) To avoid confusing weight with work, we write out the word instead of using the symbol W .Thu s , weight = mg = ( 1 . 85 × 10 4 kg )( 9 . 8m / s 2 ) 1 . 8 × 10 5 N . (b) The buoyant force is F b = ρ w gV w where ρ w =1000kg / m 3 is the density of water and V w is the volume of water displaced by the dinosaur. If we use f for the fraction of the dinosaur’s total volume V which is submerged, then V w = fV . We can further relate V to the dinosaur’s mass using the assumption that the density of the dinosaur is 90% that of water: V = m/ (0 . 9 ρ w ). Therefore, the apparent weight of the dinosaur is weight app =we ight ρ w g µ f m 0 . 9 ρ w =we ight gf m 0 . 9 . If f =0 . 50, this yields 81 kN for the apparent weight. (c) If f
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Unformatted text preview: f = 0 . 90, we Fnd the apparent weight is zero (it ﬂoats). (e) Eq. 15-8 indicates that the water pressure at that depth is greater than standard air pressure (the assumed pressure at the surface) by ρ w gh = (1000)(9 . 8)(8) = 7 . 8 × 10 4 Pa. If we assume the pressure of air in the dinosaur’s lungs is approximately standard air pressure, then this value represents the pressure di±erence which the lung muscles would have to work against. (f) Assuming the maximum pressure di±erence the muscles can work with is 8 kPa, then our previous result (78 kPa) spells doom to the wading Diplodocus hypothesis....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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