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1 year ago

a force vector f has a magnitude of 12.0 n. it is oriented 60° to the left of the y ax what are its x and y components?

1 answer:

5 0

The force vector that has a magnitude of 12.0 N. and is oriented 60° to the left of the (y) has the followings components:

v(x) =6 N
v(y) = 10.39 N

To solve this exercise the formulas and procedures we will use are:

v(x) = v * cosine (angle)
v(y) = v * sine (angle).

Where:

v= magnitude of the vector
v(x) = component of the vector on the (x) axis
v(y) = component of the vector on the (y) axis
angle = angle

Information about the problem:

angle = 60º
v = 12.0 N
v(x)= ?
v(y)= ?

Applying the formula of the component of the vector in the (x) axis we have:

v(x) = v * cosine (angle).

v(x) = 12.0 N * cosine (60º)

v(x) =6 N

Applying the formula of the component of the vector in the (y) axis we have:

v(y) = v * sine (angle)

v(y) = 12.0 N * sine (60º)

v(y) = 10.39 N

<h3>What is a vector?</h3>

It can be said to be a straight line described by a point (a) and (b) that has direction and sense.

Learn more about vector at: brainly.com/question/2094736

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Note that there's no negative sign before the fraction.

Make sure that all values are in SI units:

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$\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.20}\\&= \rm -1.2586\; J\end{aligned}$ .

Final potential energy:

$\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.90}\\&= \rm -0.279689\; J\end{aligned}$ .

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$\begin{aligned}\text{Work required} &= \text{Final EPE} - \text{Initial EPE}\\&= \rm -0.279689\; J - (-1.2586\; J)\\&\approx 0.979\; J\end{aligned}$ .

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