421_Physics ProblemsTechnical Physics

421_Physics ProblemsTechnical Physics - Chapter 14 P14.34...

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Chapter 14 423 P14.33 The balloon stops rising when ρρ air He −= bg gV Mg and air He VM , Therefore, V M e = = air He 400 125 0180 1 .. V = 1430 m 3 P14.34 Since the frog floats, the buoyant force = the weight of the frog. Also, the weight of the displaced water = weight of the frog, so ρ ooze frog Vg m g = or mV r frog ooze ooze 3 kg m m == F H G I K J × π 1 2 4 3 135 10 2 3 600 10 33 2 3 ej e j Hence, m frog kg = 0611 . . P14.35 BF g = HO sphe re sphere H O 3 glycerin sphere glycerin 2 2 kg m kg m kg m g V gV gV g V 2 1 2 500 4 10 0 10 4 500 1 250 = F H G I K J FIG. P14.35 P14.36 Constant velocity implies zero acceleration, which means that the submersible is in equilibrium under the gravitational force, the upward buoyant force, and the upward resistance force: Fm a yy 0 −× ++ + = 120 10 1100 0 4 . k g N mg w where m is the mass of the added water and
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