Physics
posted by NS .
A cargo barge is loaded in a saltwater harbor for a trip up a freshwater river.If the rectangular barge is 3.00m by 20.0m and sits 0.900m deep in the harbor, how deep will it sit in the river?

weight of freshwater displaced=LW*depth*densityfreshwater*g
weight of saltwater displaced=LW*.9*denstiy salt water*g
but the barge is afloat, so its weight is equal to the water displaced, so
the freshwater and saltwater displaced weights are equal.
LW*depth*densityfresh*g=LW*.9m*density salt*g
depth=.9 densitysalt/densityfreshwater 
I did .9(1030/1000) and i got .927...is this correct? i think it is

depends on if you believe 1030kg/m^3 is the density of surface sea water. It varies in my experience from 1020 to 1030.
If you are near a river outlet, it is on the low side. 
this piece of info was given by the prof.
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