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We know that the components of any vector along the x and y axes are given by:
[tex]\begin{gathered} v_x=v\cos\theta \\ v_y=v\sin\theta \\ \text{ where:} \\ v\text{ is the magnitude of the vector.} \\ \theta\text{ is the angle from the }x\text{ axis to the vector. } \end{gathered}[/tex]In this case we know the tha magnitude of the vector is 6.96 m/s; we also know that the angle from the x-axis to the vector is 51.5° (the sum of angles θ and the 26.5° angle); then we have:
[tex]\begin{gathered} v_x=6.96\cos51.5=4.33 \\ v_y=6.96\sin51.5=5.45 \end{gathered}[/tex]Therefore, the components of the vector are:
[tex]\begin{gathered} v_x=4.33 \\ v_y=5.45 \end{gathered}[/tex]Thanks for taking the time to read can you help me with this ap physics class from my school. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
Rewritten by : Barada
