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3 years ago

Can someone help me?!!!!!

Physics

1 answer:

saveliy_v [14]3 years ago

3 0

<h2>Hello!</h2>

The answer is:

The third option, the approximate magnitude of the given vector is 9.9 units.

$|A|=9.9units$

To calculate the magnitude (length) of a vector, we need to apply the following formula:

$|A|=\sqrt{(A_x)^{2} +(A_y)^{2} }$

So, we are given the vector:

$A=(7.6,-6.4)$

Then, substituting and calculating the magnitude of the vector, we have:

$|A|=\sqrt{(7.6)^{2} +(-6.4)^{2}}=\sqrt{57.76+40.96}=\sqrt{98.72}\\\\|A|=\sqrt{98.72}=9.9units$

Hence, we have that the correct option is the third option, the approximate magnitude of the given vector is 9.9 units.

$|A|=9.9units$

Have a nice day!

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4 years ago

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$~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s$

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The magnitude of the final velocity of Jill is given by

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