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1 answer:

6 0

I would go with C because if you do A then its all new students and thats one view D is the same. C seems like the best choice.

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Water is flowing out of a conical funnel through its apex at a rate of 12 cubic inches per minute. If the tunnel is initially fu

Lilit [14]

Check the picture below.

so, bearing in mind that, the radius and height are two sides in a right-triangle, thus both are at a ratio of each other, thus the radius is at 3:4 ratio in relation to the height.

$\bf \textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}\quad 
\begin{cases}
r=3\\
h=4
\end{cases}\implies \stackrel{full}{V}=\cfrac{\pi \cdot 3^2\cdot 4}{3}\implies \stackrel{full}{V}=12\pi 
\\\\\\
\stackrel{\frac{2}{3}~full}{V}=12\pi \cdot \cfrac{2}{3}\implies \stackrel{\frac{2}{3}~full}{V}=8\pi ~in^3
\\\\\\
\textit{is draining water at a rate of }12~\frac{in^3}{min}\qquad \cfrac{8\pi ~in^3}{12~\frac{in^3}{min}}\implies \cfrac{2\pi }{3}~min$

$\bf \textit{it takes }\frac{2\pi }{3}\textit{ minutes to drain }\frac{2}{3}\textit{ of it, leaving only }\frac{1}{3}\textit{ in it}\\\\
-------------------------------\\\\
\stackrel{\frac{1}{3}~full}{V}=12\pi \cdot \cfrac{1}{3}\implies \stackrel{\frac{1}{3}~full}{V}=4\pi\impliedby \textit{what's \underline{h} at this time?}
\\\\\\
V=\cfrac{\pi r^2 h}{3}\quad 
\begin{cases}
r=\frac{3h}{4}\\
V=4\pi 
\end{cases} \implies 4\pi =\cfrac{\pi \left( \frac{3h}{4} \right)^2 h}{3}$

$\bf 12\pi =\cfrac{\pi \cdot 3^2h^2 h}{4^2}\implies 12\pi =\cfrac{9\pi h^3}{16}\implies \cfrac{192\pi }{9\pi }=h^3
\\\\\\
\sqrt[3]{\cfrac{192\pi }{9\pi }}=h\implies \sqrt[3]{\cfrac{64}{3}}=h\implies \cfrac{4}{\sqrt[3]{3}}=h
\\\\\\
\cfrac{4}{\sqrt[3]{3}}\cdot \cfrac{\sqrt[3]{3^2}}{\sqrt[3]{3^2}}=h\implies \cfrac{4\sqrt[3]{9}}{\sqrt[3]{3^3}}=h\implies \cfrac{4\sqrt[3]{9}}{3}=h$

4 0

3 years ago

Consider the two vectors A with arrow = 3 î − ĵ and B with arrow = − î − 2 ĵ. (a) Calculate A with arrow + B with arrow 2 Correc

Verdich [7]

Answer:

Let $\vec{A}=3\hat{i}-\hat{j}$ and $\vec{B}=-\hat{i}-2\hat{j}$ .

$\vec{A}+\vec{B}=(3\hat{i}-\hat{j})+(-\hat{i}-2\hat{j})=2\hat{i}-3\hat{j}$

$\vec{A}-\vec{B}=(3\hat{i}-\hat{j})-(-\hat{i}-2\hat{j})=4\hat{i}+\hat{j}$

c) Now we calculate the direction of $\vec{A}+\vec{B}$ and $\vec{A}-\vec{B}$

We calculate the direction of $\vec{A}+\vec{B}$ like this:

$\theta=tan^{-1}(\frac{-3}{2})=-56,3^{\circ}$ . Since, $270\leq\theta\leq360$ , then the direction of $\vec{A}+\vec{B}$ is $360^{\circ}-\theta=360^{\circ}-56.3^{\circ}=303.7^{\circ}$ .