P15 084

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84. The area facing down (and up) is A = (0.050 m)(0.040 m) = 0.0020 m

2

. The submerged volume is

V = Ad where d = 0.015 m. In order to float, the downward pull of gravity mg mustequal the upward
buoyant force exerted by the seawater of density ρ:

mg = ρV g

=

⇒ m = ρV = (1025)(0.0020)(0.015) = 0.031 kg .


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