P15_059 - f = 0 . 90, we Fnd the apparent weight is zero...

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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 oats). (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 w gh = (1000)(9 . 8)(8) = 7 . 8 10 4 Pa. If we assume the pressure of air in the dinosaurs lungs is approximately standard air pressure, then this value represents the pressure dierence which the lung muscles would have to work against. (f) Assuming the maximum pressure dierence 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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