step by step it: Fluid Mechanics Name Block: Buoyancy 0 07 m long and 0.09 n tal

Question

step by step

it: Fluid Mechanics Name Block: Buoyancy 0 07 m long and 0.09 n tall and has a mas of 050 kg. The block is p the dessity of water at 25e 1. A block is 012 m wide, loating in water at a temperature of 25 C. (You ean look u in your Physics Reference Tables.) (a) If theblock weighs 0.50 kg, what volume of the block will be below the surface of the water? 10- m b) If the entire block were pushed under water, bow much water would it displace? c) How much additional mas could be piled on top of the block before it sinks?

Explanation / Answer

(A) volume, V = 0.12 x 0.07 x 0.09 = 7.56 x 10^-4 m^3

buoyant force, Fb = V_sub rho_water g

for equilibrium,

Fb = m g

(V_sub)(997)(g) = (0.50)(g)

V_sub = 5 x 10^-4 m^3

volume below water = 5 x 10^-4 m^3

(B) V = 0.12 x 0.09 x 0.07 = 7.56 x 10^-4 m^3

(C) Fb' = (7.56 x 10^-4)(997)(g)

FB' = 0.75g

delta(m) = 0.75 - 0.50 = 0.25 kg

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