[填空题]
$\text { If } V_{x}=6.80 \text { units and } V_{y}=-7.40 \text { units, determine the magnitude and direction of } \overrightarrow{\mathbf{V}} \text {. }$ magnitude units direction $^{\circ}$
参考答案: 空格1: 10±2%
空格2: -47±2%
本题详细解析: Given that $V_{x}=6.80$ units and $ V_{y}=-7.40 $ units, the magnitude of $\overrightarrow{\mathbf{V}}$ is given by $ V=\sqrt{V_{x}^{2}+V_{y}^{2}}=\sqrt{6.80^{2}+(-7.40)^{2}}=10.0$ units. The direction is given by an angle of $ \theta=\tan ^{-1} \frac{-7.40}{6.80}=-47^{\circ}$ , or $47^{\circ}$ below the positive x axis.
