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(B) The angular momentum $\overrightarrow{L}$ is given by the cross product of the position vector $\overrightarrow{r}$ and the linear momentum $\over

Vedclass · Accepted Answer

$91$

Vedclass · Answer

(B) The angular momentum $\overrightarrow{L}$ is given by the cross product of the position vector $\overrightarrow{r}$ and the linear momentum $\overrightarrow{p} = m\overrightarrow{v}$.Given $m = 1\,kg$,$\overrightarrow{r} = (3\hat{i} - \hat{j})\,m$,and $\overrightarrow{v} = (3\hat{j} + \hat{k})\,m/s$.$\overrightarrow{L} = \overrightarrow{r} 	imes (m\overrightarrow{v}) = 1 \cdot [(3\hat{i} - \hat{j}) 	imes (3\hat{j} + \hat{k})]$.Calculating the cross product:$\overrightarrow{L} = 3\hat{i} 	imes 3\hat{j} + 3\hat{i} 	imes \hat{k} - \hat{j} 	imes 3\hat{j} - \hat{j} 	imes \hat{k}$.Using unit vector properties ($\hat{i} 	imes \hat{j} = \hat{k}$,$\hat{i} 	imes \hat{k} = -\hat{j}$,$\hat{j} 	imes \hat{j} = 0$,$\hat{j} 	imes \hat{k} = \hat{i}$):$\overrightarrow{L} = 9\hat{k} - 3\hat{j} - 0 - \hat{i} = -\hat{i} - 3\hat{j} + 9\hat{k}$.The magnitude is $|\overrightarrow{L}| = \sqrt{(-1)^2 + (-3)^2 + (9)^2} = \sqrt{1 + 9 + 81} = \sqrt{91}$.Thus,$x = 91$.

The position vector of a $1\,kg$ object is $\overrightarrow{r} = (3\hat{i} - \hat{j})\,m$ and its velocity is $\overrightarrow{v} = (3\hat{j} + \hat{k})\,m/s$. The magnitude of its angular momentum is $\sqrt{x}\,N\cdot m\cdot s$,where $x$ is:

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