1. The problem statement, all variables and given/known data 1. A block is 7 cm wide, 8 cm long, and 2 cm tall and has a mass of 67 g. a) What is its volume? b) What is its density? c) Floating in water (with the 7x8 face down), how deep will the bottom be (what is x in figure 2)? d) How much extra weight can it support without sinking? 2. A 2 cm by 2 cm by 2 cm metal block has a mass of 80 g. a) What volume of water does it displace when it is submerged? b) What is the weight of the water displaced? c) What is the apparent weight of the block when measured under water? 2. Relevant equations A=Lw. V=Ah=Lwh. density = m/v Fbuoyancy = (Mass displaced water)*g mblock*g = Fbuoyancy =mdisplaced water*g ---->density of block* volume of block = density water* Volume displaced water ----> DENSITY BLOCK*H = DENSITY WATER X Fbouyancy = density water*Volume of block 3. The attempt at a solution 1a)V=Ah=Lwh. (.08m)(.07m)(.02m) V=0.000112m^3 1b)density = m/v .067kg/0.000112m^3 = 598.214kg/m^3 c) Density block* h = densitywater*x (598.214kg/m^3)(.02m) =(1000kg/m^3) x x=0.019964m d)m displaced water = Vblock* density water Mdw = (.000112m^3)(1000kg/m^3) Mdw = 0.112kg Mblock = .067kg mass block - massdw = .045kg 2a) V=Lwh V=(.02)(.02)(.02) = .000008m^3 Density = 10,000kg/m^3 Pbmetal*Vmetal = density water*Vdisplaced water (10,000kg/m^3)(.000008m^3) = (1000kg/m^3)( Vdw) Vdw = 0.00008m^3 b)F bouyancy = density water * vblock*g =(1000kg/m^3)(.000008m^3)(9.8m/s^2) =0.0784N dw =(1000kg/^3)(.000008m^3) =.008kg Fbouyancy = Mdw*g= .0784N c) I dont know... F bouyancy = density water * vblock*g =(1000kg/m^3)(.000008m^3)(9.8m/s^2) =0.0784N Can someone check my work? Question 2a-c, Im unsure of. I need some explanation or some ideas to solve 2b and 2c.