Category: Vector

Question Number 16502 by Tinkutara last updated on 23/Jun/17 $$\mathrm{The}\mathrm{acceleration}\mathrm{vector}\mathrm{of}\mathrm{a}\mathrm{particle}\mathrm{in}$$$$\mathrm{uniform}\mathrm{circular}\mathrm{motion}\mathrm{averaged}\mathrm{over}$$$$\mathrm{one}\mathrm{cycle}\mathrm{is}\mathrm{a}\mathrm{null}\mathrm{vector}.\left(\mathrm{True}/\mathrm{False}\right)$$ Answered by ajfour last updated on 23/Jun/17 $$\bar {{a}}=\:−\frac{{v}_{\mathrm{0}}…

Question Number 16401 by sushmitak last updated on 21/Jun/17 $$\mathit{v}=2\mathit{i}+2\mathit{j}+5\mathit{k}$$$$\mathit{r}=\mathit{i}+9\mathit{j}-8\mathit{k}$$$$\mathrm{Find}\mathit{\omega }$$$$\mathrm{I}\mathrm{can}\mathrm{do}\frac{\mathit{r}\times \mathit{v}}{r^2}=\mathit{\omega }$$$$\mathrm{and}\mathrm{i}\mathrm{get}\mathit{\omega }=\frac{61\mathit{i}-21\mathit{j}-16\mathit{k}}{146}$$$$\mathrm{but}\mathrm{i}\mathrm{dont}\mathrm{get}\mathit{w}\times \mathit{r}=\mathit{v}.$$$$why?$$…

Question Number 16294 by Tinkutara last updated on 20/Jun/17 $$\mathrm{Two}\mathrm{particles},1\mathrm{and}2,\mathrm{move}\mathrm{with}$$$$\mathrm{constant}\mathrm{velocities}\overrightarrow{v_1}\mathrm{and}\overrightarrow{v_2}.\mathrm{At}\mathrm{the}$$$$\mathrm{initial}\mathrm{moment},\mathrm{their}\mathrm{position}\mathrm{vectors}$$$$\mathrm{are}\:\mathrm{equal}\:\mathrm{to}\:\overset{\rightarrow} {{r}_{\mathrm{1}} }\:\mathrm{and}\:\overset{\rightarrow} {{r}_{\mathrm{2}} }.\:\mathrm{How}\:\mathrm{must}\:\mathrm{these} \…

Question Number 16153 by Tinkutara last updated on 18/Jun/17 $$\mathrm{Is}\mathrm{angular}\mathrm{displacement}\mathrm{a}\mathrm{vector}$$$$\mathrm{quantity}?$$ Answered by ajfour last updated on 18/Jun/17 $$onlyaslongasitisverysmall.$$ Commented…

Question Number 147162 by nadovic last updated on 18/Jul/21 Answered by liberty last updated on 18/Jul/21 $$\left(1\right)\overrightarrow{p}=\frac{2}{5}\overrightarrow{a}+\frac{1}{5}\overrightarrow{b}+\frac{2}{5}\overrightarrow{c}$$$$\left(\mathrm{2}\right)\overset{\rightarrow} {{p}}.\overset{\rightarrow} {{b}}=\overset{\rightarrow}…