A raft has a surface area of 100m^2 and a height of 10.0cm. If it is constructed

Question

A raft has a surface area of 100m^2 and a height of 10.0cm. If it is constructed of yellow pine and floating in a lake, calculate how much of the raft will be above the waterline

Explanation / Answer

let (h) meter be the height [ABOVE] the waterline >> (0.10 - h) height is below the waterline >>> volume submerged (V) = A * (0.10 - h) where A = area of crossection = 100 m^2 V = 100 (0.10 - h) -------------------------------- Archi. principle >> weight of submerged object = weight of water displaced mass * g = V *density (water) *g mass = 100 (0.10 - h) *d (water) density (wood)*(total wood volume ) =100(0.10 - h) *d (water) d(wood) * (100*0.10) = 100(0.10 - h) *d (water) d(wood) = 10(0.10 - h) *d (water) ===================== d(water) =1000 kg/m^3 ---------------------- d(wood) = 10000(0.10 - h) = 1000 - 10000 h 10000 h = 1000 - d(wood) h = [1000 - d(wood)] /10000 h = [0.10 - {d(wood)/10000}] meter I do not know the density of wood >>>> put & calculate convert to centimeter>> =============================== 2) let (n) persons of mass (m=70 kg) are placed on raft so that full (0.10 m) height of wood goes inside the waterline total weight = water displaced (full volume in) n*mg + d(wood)*(100*0.10) g = [100*0.10] d(water) *g n*70 + d(wood)*(10) = [10] d(water) calculate (n) > if n = 8.56 (fraction) then only n=8 will be the answer >> because people can't be in fraction, and n =9 will make the raft SINK.

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