Question Number 61823 by necx1 last updated on 09/Jun/19 $${If}\:{D},{E}\:{and}\:{F}\:{are}\:{midpoints}\:{of}\:{the}\:{sides} \\ $$$${BC},{CA}\:{and}\:{AB}\:{respectively}\:{of}\:{the}\:\bigtriangleup{ABC} \\ $$$${and}\:{O}\:{be}\:{any}\:{point}.{Prove}\:{that} \\ $$$${O}\overset{\rightarrow} {{A}}\:+\:{O}\overset{\rightarrow} {{B}}\:+{O}\overset{\rightarrow} {{C}}={O}\overset{\rightarrow} {{D}}+{O}\overset{\rightarrow} {{E}}+{O}\overset{\rightarrow} {{F}} \\ $$ Answered…
Question Number 126786 by Ar Brandon last updated on 24/Dec/20 Answered by Olaf last updated on 26/Dec/20 $$\mathrm{C}_{\mathrm{1}} \:=\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\left(\mathrm{first}\:\mathrm{city}\right) \\ $$$$\mathrm{C}_{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{60}\sqrt{\mathrm{2}}−\mathrm{20}}\\{\mathrm{75}+\mathrm{60}\sqrt{\mathrm{2}}}\end{pmatrix}\:\left(\mathrm{second}\:\mathrm{city}\right) \\ $$$$\left(\mathrm{C}_{\mathrm{1}} \mathrm{C}_{\mathrm{2}}…
Question Number 126778 by Ar Brandon last updated on 24/Dec/20 Answered by Dwaipayan Shikari last updated on 24/Dec/20 $$\sqrt{\left(\overset{\rightarrow} {{A}}_{\mathrm{1}} \right)^{\mathrm{2}} +\left(\overset{\rightarrow} {{A}}_{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}\mid\overset{\rightarrow}…
Question Number 126716 by sdfg last updated on 23/Dec/20 Answered by liberty last updated on 23/Dec/20 $$\overset{\rightarrow} {{a}}=\left(\mathrm{3},\mathrm{4}\right)\:;\:\overset{\rightarrow} {{b}}=\left({x},{y}\right)\:\rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{5}…\left({i}\right) \\ $$$$\:\overset{\rightarrow} {{a}}\:\bot\:\overset{\rightarrow} {{b}}\:\rightarrow\:\overset{\rightarrow}…
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Question Number 126567 by sdfg last updated on 21/Dec/20 Answered by physicstutes last updated on 22/Dec/20 $$\:\boldsymbol{\mathrm{A}}\:=\:\left({k},{k}\right)\:\mathrm{and}\:\boldsymbol{\mathrm{B}}\:=\:\left(−\mathrm{3},\mathrm{4}\right) \\ $$$$\boldsymbol{\mathrm{A}}.\boldsymbol{\mathrm{B}}\:=\:\mid\boldsymbol{\mathrm{A}}\mid\mid\boldsymbol{\mathrm{B}}\mid\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:−\mathrm{3}{k}+\mathrm{4}{k}\:=\:{k} \\ $$$$\Rightarrow\:{k}\:=\:\sqrt{{k}^{\mathrm{2}} +{k}^{\mathrm{2}} }\:.\sqrt{\left(−\mathrm{3}\right)^{\mathrm{2}}…
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Question Number 126475 by sdfg last updated on 20/Dec/20 Answered by physicstutes last updated on 20/Dec/20 $$\:\left(\mathrm{a}\right)\:\boldsymbol{\mathrm{a}}\:=\:\frac{\boldsymbol{\Delta\mathrm{v}}}{\boldsymbol{\Delta}{t}}\:=\:\frac{\left[\left(\mathrm{10}\:\boldsymbol{\mathrm{j}}\right)\:−\left(\mathrm{10}\:\boldsymbol{\mathrm{i}}\:+\:\mathrm{20}\:\boldsymbol{\mathrm{j}}\right)\right]\:\mathrm{m}\:\mathrm{s}^{−\mathrm{1}} }{\left(\mathrm{4}−\mathrm{0}\right)\:\mathrm{s}}\:=\:\left[−\left(\frac{\mathrm{10}}{\mathrm{4}}\right)\:\boldsymbol{\mathrm{i}}\:−\:\left(\frac{\mathrm{10}}{\mathrm{4}}\right)\:\boldsymbol{\mathrm{j}}\:\right]\mathrm{m}\:\mathrm{s}^{−\mathrm{2}} \\ $$$$\mathrm{or}\:\boldsymbol{\mathrm{a}}\:=\left(\:−\frac{\mathrm{5}}{\mathrm{2}}\boldsymbol{\mathrm{i}}\:−\:\frac{\mathrm{5}}{\mathrm{2}}\boldsymbol{\mathrm{j}}\:\right)\:\mathrm{m}\:\mathrm{s}^{−\mathrm{2}} \:\:\:\mathrm{ofcourse}\:\mathrm{assuming}\:\mathrm{we}\:\mathrm{are}\:\mathrm{using}\:\mathrm{SI}\:\mathrm{base}\:\mathrm{units}. \\ $$$$\:\left(\mathrm{b}\right)\:\boldsymbol{\mathrm{r}}\:=\:\boldsymbol{\mathrm{r}}_{\mathrm{0}} +\:\:\boldsymbol{\mathrm{v}}_{\mathrm{0}} {t}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{a}}\:{t}^{\mathrm{2}}…