Asked by Anonymous
HOw many cubic meter of water will an air tight rectangle box with a mass of 8000 kg and dimensions of 10m long 5m wide 3m tall displace when dropped in water? will it float?
Answers
Answered by
Henry
V = L*W*h = 10 * 5 8 3 ,
= 150 m^3 = 1.50*10^8 cm^3 = Vol. of box.
Db = 8*10^6g * (1/1.5*10^8cm^3) = 0.0533 g/cm^3 . = Density ofbox.
Dw = 1g/cm^3 = Density of water.
Vb = (Db/Dw)*V,
Vb = (0.0533 / 1)*150 = 8 m^3 = Vol.
below water = Vol. of water displaced.
The box will float, because its' density is less than density of water.
= 150 m^3 = 1.50*10^8 cm^3 = Vol. of box.
Db = 8*10^6g * (1/1.5*10^8cm^3) = 0.0533 g/cm^3 . = Density ofbox.
Dw = 1g/cm^3 = Density of water.
Vb = (Db/Dw)*V,
Vb = (0.0533 / 1)*150 = 8 m^3 = Vol.
below water = Vol. of water displaced.
The box will float, because its' density is less than density of water.
Answered by
Henry
Correction: V = 10 * 5 * 3 = 150 m^3.
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