Notes on Chapter 3 examples

B15 agg w vg disp liq b b 2 ag g 100 g cm3 b expt

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Buoyant force up = Weight of block down Wdisplaced liquid oil 2 soln B 0.75 bSGg 15 g / cm3 . B soln A w 15 . 3.9 Let Volume displaced: Vd 1 hs 1 WA + WB h hb1 for b p 1 hb 1 Vw Ap h p1 bg subst. 3 for h p 1 in b 2 g, solve for h b1 hb1 i d Subst. (1) for Vd 1 , solve for h p1 hb1 i (2) Ap h p1 Vd 1 WA WB pw g Vw Ap bW A h p1 WB pw g bi g Vw Ap g LM 1 NM A p 3-3 Vw Ap h p1 WA WB pw gAp 1 Ab OP QP (1) WA WB weight of displaced water Volume of pond water: Vw subst. 2 d (object sinks) Ab hp1 hb1 wVd 1 g WA WB pw gAb h p1 hb1 Ab hsi Archimedes 1 Before object is jettisoned bg density of water. Note: w d Ab h p1 hb1 (3) (4) i...
View Full Document

Ask a homework question - tutors are online