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anonymous
 4 years ago
Find the apparent weight of an object fully submerged in water if its volume is 2m^3 and its density is 4000kg/m^3.
anonymous
 4 years ago
Find the apparent weight of an object fully submerged in water if its volume is 2m^3 and its density is 4000kg/m^3.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I know that apparent weight is the weight that the scale has to support. Also I know there are three forces acting upon the object: buoyancy and tension frm the scale going upwards while the weight is going downwards.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am just really confused about what equations to use

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(Gravitionnal) weight = P = mg = Vdg = 2*4000*10=80 000 N

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Buoyant force = Vd(water)g=2*1000*10=20 000 N

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So the apparent weight is P(Buoyant Force) = 80 000  20 000 = 60 000 N

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so for buoyant force it's always referring to the use of water's density and not the object's density?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes indeed. Par le principe d'Archimède, ''Tout corps plongé dans un fluide subit de la part de ce fluide une poussée verticale, orientée vers le haut, dont la valeur est égale au poids du volume de fluide déplacé''.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A body immersed in a fluid receives from the fluid a vertical thrust, facing upward, whose magnitude equals the weight of the displaced fluid.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In this case, the fluid is water, so the the buoyancy thrust involves the density of water :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh that makes sense. Thank you so much! ^_^
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