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A certain vector in the $x y$-plane has an $x$-component of $4 \mathrm{~m}$ and a $y$-component of $10 \mathrm{~m}$. It is then rotated in the $x y$-plane so that its $x$-component is doubled. Then its new $y$-component is (approximately)
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1319 Upvotes
Verified Answer
The correct answer is:
$7.2 \mathrm{~m}$
Here, $\mathbf{A}=4 \mathbf{i}+10 \mathbf{j}$
$\begin{aligned}
\therefore \quad|\mathrm{A}|=\sqrt{(4)^2+(10)^2} & =\sqrt{16+100} \\
& =\sqrt{116} \mathrm{~m}
\end{aligned}$
Now, according to question,
So,
$\begin{aligned}
\mathbf{A} & =8 \mathbf{i}+n \mathbf{j} \\
|\mathbf{A}| & =\sqrt{(8)^2+n^2} \\
\sqrt{116} & =\sqrt{(8)^2+n^2}
\end{aligned}$
squaring both sides
$\begin{aligned}
116 & =64+n^2 \\
or \quad n^2 & =116-64=52 \\
or \quad n & =7.2 \mathrm{~m}
\end{aligned}$
$\begin{aligned}
\therefore \quad|\mathrm{A}|=\sqrt{(4)^2+(10)^2} & =\sqrt{16+100} \\
& =\sqrt{116} \mathrm{~m}
\end{aligned}$
Now, according to question,
So,
$\begin{aligned}
\mathbf{A} & =8 \mathbf{i}+n \mathbf{j} \\
|\mathbf{A}| & =\sqrt{(8)^2+n^2} \\
\sqrt{116} & =\sqrt{(8)^2+n^2}
\end{aligned}$
squaring both sides
$\begin{aligned}
116 & =64+n^2 \\
or \quad n^2 & =116-64=52 \\
or \quad n & =7.2 \mathrm{~m}
\end{aligned}$
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