Respuesta :
[tex]\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
F=\textit{drag force}\\
w=\textit{wet area}\\
s=speed
\end{cases}\textit{\underline{F} varies jointly with \underline{w} and square of \underline{s}}\quad F=kws^2[/tex]
[tex]\bf \textit{we also know that } \begin{cases} w=502\\ F=98\\ s=7 \end{cases}\implies 98=k502\cdot 7^2 \\\\\\ 98=24598k \implies \cfrac{98}{24598}=k\implies \cfrac{1}{251}=k\qquad \implies \boxed{F=\cfrac{1}{251}ws^2}\\\\ -------------------------------\\\\ \textit{now, what's \underline{w} when } \begin{cases} s=7.5\\ F=135 \end{cases}\implies 135=\cfrac{1}{251}w7.5^2 \\\\\\ \cfrac{135\cdot 251}{56.25}=w\implies 604.2=w[/tex]
[tex]\bf \textit{we also know that } \begin{cases} w=502\\ F=98\\ s=7 \end{cases}\implies 98=k502\cdot 7^2 \\\\\\ 98=24598k \implies \cfrac{98}{24598}=k\implies \cfrac{1}{251}=k\qquad \implies \boxed{F=\cfrac{1}{251}ws^2}\\\\ -------------------------------\\\\ \textit{now, what's \underline{w} when } \begin{cases} s=7.5\\ F=135 \end{cases}\implies 135=\cfrac{1}{251}w7.5^2 \\\\\\ \cfrac{135\cdot 251}{56.25}=w\implies 604.2=w[/tex]
Variation can be direct, inverse or joint.
The wet surface area is 60 square feet
The variation is given as:
[tex]\mathbf{F\ \alpha\ As^2}[/tex]
Express as an equation
[tex]\mathbf{F\ \ =\ kAs^2}[/tex]
Make k the subject.
[tex]\mathbf{k = \frac{F}{As^2}}[/tex]
Rewrite as:
[tex]\mathbf{ \frac{F_1}{A_1s_1^2} = \frac{F_2}{A_2s_2^2}}[/tex]
Where:
[tex]\mathbf{F_1 = 98,\ A_1 = 50,\ s_1 = 7}[/tex]
[tex]\mathbf{F_2 = 135,\ s_2 = 7.5}[/tex]
So, we have:
[tex]\mathbf{ \frac{98}{50 \times 7^2} = \frac{135}{A_2 \times 7.5^2}}[/tex]
[tex]\mathbf{ \frac{98}{50 \times 49} = \frac{135}{A_2 \times 56.25}}[/tex]
[tex]\mathbf{ \frac{2}{50} = \frac{135}{A_2 \times 56.25}}[/tex]
[tex]\mathbf{ \frac{1}{25} = \frac{135}{A_2 \times 56.25}}[/tex]
Make A2 the subject
[tex]\mathbf{A_2 = \frac{25 \times 135}{56.25}}[/tex]
[tex]\mathbf{A_2 = \frac{3375}{56.25}}[/tex]
[tex]\mathbf{A_2 = 60}[/tex]
Hence, the wet surface area is 60 square feet
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