## Related questions with answers

Question

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

$d_s$

is placed in a liquid of density

$d_L=62.5lb/ft^3$

,it will sink to a depth h where .

$\dfrac{π}{3}(3rh^2-h^3)d_L=\dfrac{4}{3}πr^3d_s$

Find an approximate value for h if:

$r=5 ft$

and

$d_s=20lb/ft^3$

Solution

VerifiedStep 1

1 of 3We substitute $r$, $d_L$ and $d_S$ as indicated:

$\frac{\pi}{3}\left(3(5)h^2-h^3\right)62.5 = \frac{4}{3}\pi \left(5^3\right)(45)$

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

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

Simplify:

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

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

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

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