In both cases shown in the diagram below, a block is floating at rest in a liquid. The liquids in the two beakers have the same density, that is, PA = PB. Both blocks have 55% of their volume below the surface, but the volumes of the blocks are not the same. Find the mass m2 in terms of V1, V2, and m1. Block 1 Block 2 m1 m2 РА PB V m2 = m1 V2 V2 m2 = m1 (V2–V1)' (V2+V1). m2 = mị тр — 0.55m О то — 0.45m
In both cases shown in the diagram below, a block is floating at rest in a liquid. The liquids in the two beakers have the same density, that is, PA = PB. Both blocks have 55% of their volume below the surface, but the volumes of the blocks are not the same. Find the mass m2 in terms of V1, V2, and m1. Block 1 Block 2 m1 m2 РА PB V m2 = m1 V2 V2 m2 = m1 (V2–V1)' (V2+V1). m2 = mị тр — 0.55m О то — 0.45m
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Transcribed Image Text:In both cases shown in the diagram below, a block is floating at rest in a liquid. The liquids in the two beakers have the same
density, that is, PA = PB. Both blocks have 55% of their volume below the surface, but the volumes of the blocks are not the
same.
Find the mass m2 in terms of V1, V2, and m1.
Block 1
Block 2
mi
m2
РА
Рв
Vị
m2 = mị
V,
V2
m2 = mj
Vị
(V2–V1)
(V2+V1),
m2 = mj
m2 = 0.55mị
m2 = 0.45mị
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