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Calculate the buoyant force acting on it if a stone of mass 250g is thrown in water?​

User Clonejo
by
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1 Answer

3 votes

Answer:


F = 1.114\,N

Step-by-step explanation:

A common stone may have densities between
1600\,(kg)/(m^(3)) and
3300\,(kg)/(m^(3)). Let consider a rock with a density of
2200\,(kg)/(m^(2)). The buoyant force acting on the stone has the following formula:


F = (\rho_(w))/(\rho_(s))\cdot m\cdot g


F = (1000\,(kg)/(m^(3)) )/(2200\,(kg)/(m^(3)) ) \cdot (0.25\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right)


F = 1.114\,N

User Gunas
by
8.5k points

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