A 1.60-kg particle moves in the xy plane with a velocity of v with arrow = (4.60 î − 3.80 ĵ) m/s. Determine the angular momentum of the particle about the origin when its position vector is r with arrow = (1.50 î + 2.20 ĵ) m.
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Step 1
The cross product of two vectors A and B is given by the formula A x B = |A| |B| sin(theta) n, where |A| and |B| are the magnitudes of the vectors, theta is the angle between the vectors, and n is a unit vector perpendicular to both A and B. In this case, the Show more…
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A 1.60-kg particle moves in the xy plane with a velocity of v with arrow = (4.00 i - 4.10 j) m/s. Determine the angular momentum of the particle about the origin when its position vector is r with arrow = (1.50 i + 2.20 j) m.
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A 1.50 -kg particle moves in the $x y$ plane with a velocity of $\overrightarrow{\mathbf{v}}=(4.20 \mathbf{i}-3.60 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} .$ Determine the angular momentum of the particle about the origin when its position vector is $\overrightarrow{\mathbf{r}}=(1.50 \mathbf{i}+2.20 \hat{\mathbf{j}}) \mathrm{m} .$
A 1.50 -kg particle moves in the $x$ y plane with a velocity of $\overrightarrow{\mathbf{v}}=(4.20 \hat{\mathbf{i}}-3.60 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} .$ Determine the angular momentum of the particle about the origin when its position vector is $\overrightarrow{\mathbf{r}}=(1.50 \hat{\mathbf{i}}+2.20 \hat{\mathbf{j}}) \mathrm{m}$.
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