Answer to Question #241437 in Classical Mechanics for obi

Question #241437

How much water (density = 1.00 X 103 kg/m3 ) must be displaced to float a cubic block of wood (density 655 kg/m3) that is 15.0 cm on a side?

1
Expert's answer
2021-09-24T09:25:57-0400

a=15cm=0.15ma = 15cm = 0.15m

ρw=1103kgm3\rho_w = 1*10^3\frac{kg}{m^3}

ρc=655kgm3\rho _c= 655\frac{kg}{m^3}

g=9.8kgm3g = 9.8 \frac{kg}{m^3}

Forces acting on a wooden cube:\text{Forces acting on a wooden cube:}

FA=ρwgV1F_A = \rho_wgV_1

V1The volume of a part of the cube immersed inV_1 - \text{The volume of a part of the cube immersed in}

water and also the volume of displaced water\text{water and also the volume of displaced water}

Fg=mcgF_g = m_cg

mcmass of a wooden cubem_c- \text{mass of a wooden cube}

mc=Vcρc=a3ρcm_c = V_c\rho_c = a^3\rho_c

Fg=FAF_g = F_A

a3ρcg=ρwgV1a^3\rho_c g = \rho_wgV_1

V1=a3ρcρwg=0.153665103=2.24103m3V_1 = \frac{a^3\rho_c} {\rho_wg}= \frac{0.15^3*665}{10^3}=2.24*10^{-3}m^3

m1mass of displaced waterm_1 -\text{mass of displaced water}

m1=ρwV1=2.24kgm_1 = \rho_wV_1= 2.24kg

Answer:\text{Answer:}

2.24kgmass of displaced water2.24kg-\text{mass of displaced water}

2.24103m3-volume of displaced water2.24*10^{-3}m^3 \text{-volume of displaced water}


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