(a) mass of lead needed if the leads is on top..
ml=mass of lead ?
for this part the amount immersed is wood
Fb=?Vg
the force displaced is equal to the weight
Fb=Fg
?wVg(mw/?wood)=(mw+ml)g,gravity cancels out and solve for ml
ml=?wV(mw/?wood)-mw=
plug and solve
(b)this time the lead and block are immersed, the block holdsthe lead.
?wVg(mw/?wood)+(?w/?l)mlg=(mw+ml)g, gravity cancels out, and solve for ml
ml=?wV(mw/?wood)+(?w/?l)ml-mw
ml=pwv(mw/?wood)-mw=
plug and solve
(?w/?l)
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