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If the water displaced by an object has a volume of 0.1 m3, what is the buoyant force exerted

on that object?

1 Answer

6 votes

Answer:

Approximately
1000 \; {\rm N} (assuming that
g = 10\; {\rm N \cdot kg^(-1)}.)

Step-by-step explanation:

The buoyant force on an object is equal to the weight of the liquid that this object has displaced.

In this example, the object displaced
V = 0.1\; {\rm m^(3)} of water. The density of water is
\rho = 1.00 * 10^(3)\; {\rm kg \cdot m^(-3)}. Thus, the mass of the water displaced would be
m = \rho\, V = 1.00 * 10^(2)\; {\rm kg}.

Since
g = 10\; {\rm N \cdot kg^(-1)} by assumption, the weight of that
m = 1.00 * 10^(2)\; {\rm kg} of water would be
m\, g = 1.00 * 10^(3)\; {\rm N}. Hence, the buoyant force on this object would be
1.00 * 10^(3)\; {\rm N}, which is
1000\; {\rm N} when rounded to one significant figure (as in volume.)

User Jelissa
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