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Question

use Archimedes’ Principle, which states that when a sphere of radius r with density

dsd_s

is placed in a liquid of density

dL=62.5lb/ft3d_L=62.5lb/ft^3

,it will sink to a depth h where .

π3(3rh2h3)dL=43πr3ds\dfrac{π}{3}(3rh^2-h^3)d_L=\dfrac{4}{3}πr^3d_s

Find an approximate value for h if:

r=5ftr=5 ft

and

ds=20lb/ft3d_s=20lb/ft^3

Solution

Verified
Step 1
1 of 3

We substitute rr, dLd_L and dSd_S as indicated:

π3(3(5)h2h3)62.5=43π(53)(45)\frac{\pi}{3}\left(3(5)h^2-h^3\right)62.5 = \frac{4}{3}\pi \left(5^3\right)(45)

Multiply both sides by 3π\frac{3}{\pi}

(3(5)h2h3)62.5=4(53)(20)\left(3(5)h^2-h^3\right)62.5 = 4 \left(5^3\right)(20)

Simplify:

(15h2h3)62.5=22500h3+15h2360=0\begin{gather*} \left(15h^2-h^3\right)62.5 = 22500\\ -h^3+15h^2-360 = 0 \end{gather*}

We graph the function and find that it has three real zeros at approximately

h{4.32,6.51,12.80}h \in \left\{-4.32, 6.51, 12.80\right\}

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