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A certain vector in the $x y$-plane has an $x$-component of 4 m and a $y$-component of 10 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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Verified Answer
The correct answer is:
7.2 m
Here, $A=4 \mathbf{i}+10 \mathbf{j}$
$\therefore|A|=\sqrt{(4)^2+(10)^2}=\sqrt{16+100}$
$=\sqrt{116} \mathrm{~m}$
Now, according to question,
$A=8 \mathbf{i}+n \mathbf{j}$
So, $\quad|A|=\sqrt{(8)^2+n^2}$
$\therefore \quad \sqrt{116}=\sqrt{(8)^2+n^2}$
Squaring on both sides
$116=64+n^2$
or $n^2=116-64=52$
or $n=7.2 \mathrm{~m}$
$\therefore|A|=\sqrt{(4)^2+(10)^2}=\sqrt{16+100}$
$=\sqrt{116} \mathrm{~m}$
Now, according to question,
$A=8 \mathbf{i}+n \mathbf{j}$
So, $\quad|A|=\sqrt{(8)^2+n^2}$
$\therefore \quad \sqrt{116}=\sqrt{(8)^2+n^2}$
Squaring on both sides
$116=64+n^2$
or $n^2=116-64=52$
or $n=7.2 \mathrm{~m}$
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