Float boya

A 0.5 meter spherical float is used as a location mark for a fishing boat anchor. It floats in salt water. Find the depth to which the float sinks if the material of which the float is made weighs 8 kilograms per cubic meter and salt water weighs 1027 kg/m3.

Correct result:

h =  0.045 m

Solution:

R=0.5 m1=m2 8 4/3πR3=1027/3πh2(4Rh) 8 4/3π0.53=1027/3πh2(4 0.5h) 4.18879=(2150.941075.47 h)h2 h1=0.04365570.0437 h2=0.04463040.0446 h3=1.999031.999  0<h<2R h=0.0446304=0.045 mR=0.5 \ \\ m_{ 1 }=m_{ 2 } \ \\ 8 \cdot \ 4/3 \pi R^3=1027 / 3 \pi h^2(4R-h) \ \\ 8 \cdot \ 4/3 \pi 0.5^3=1027 /3 \pi h^2(4 \cdot \ 0.5 -h) \ \\ 4.18879=(2150.94 - 1075.47 \ h) h^2 \ \\ h_{1}=-0.0436557 \doteq -0.0437 \ \\ h_{2}=0.0446304 \doteq 0.0446 \ \\ h_{3}=1.99903 \doteq 1.999 \ \\ \ \\ 0 < h < 2R \ \\ h=0.0446304=0.045 \ \text{m}



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