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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}} \\ $$$$\mathrm{or}\:\boldsymbol{\mathrm{a}}\:=\left(\:−\frac{\mathrm{5}}{\mathrm{2}}\boldsymbol{\mathrm{i}}\:−\:\frac{\mathrm{5}}{\mathrm{2}}\boldsymbol{\mathrm{j}}\:\right)\:\mathrm{m}\:\mathrm{s}^{−\mathrm{2}} \:\:\:\mathrm{ofcourse}\:\mathrm{assuming}\:\mathrm{we}\:\mathrm{are}\:\mathrm{using}\:\mathrm{SI}\:\mathrm{base}\:\mathrm{units}. \\ $$$$\:\left(\mathrm{b}\right)\:\boldsymbol{\mathrm{r}}\:=\:\boldsymbol{\mathrm{r}}_{\mathrm{0}} +\:\:\boldsymbol{\mathrm{v}}_{\mathrm{0}} {t}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{a}}\:{t}^{\mathrm{2}}…
Question Number 126442 by BHOOPENDRA last updated on 20/Dec/20 $${maximum}\:{value}\:{of}\:{direction}\:{derivative}\:{of}\:\Phi:{e}^{\mathrm{3}{x}} {sin}\left({yz}^{\mathrm{4}} \right){at}\left(\mathrm{0}\:\:,\pi/\mathrm{4},\:\:\mathrm{1}\right)\:{is}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126414 by sdfg last updated on 20/Dec/20 Answered by physicstutes last updated on 20/Dec/20 $$\:\mathrm{Q}_{\mathrm{2}} .\:\mathrm{normally}\:\mathrm{we}\:\mathrm{should}\:\mathrm{be}\:\mathrm{computing}\:\mid\boldsymbol{\mathrm{A}}\:×\:\boldsymbol{\mathrm{B}}\mid\: \\ $$$$\:\mathrm{the}\:\mathrm{unit}\:\mathrm{vector}\:\boldsymbol{\mathrm{n}}\:\mathrm{could}\:\mathrm{be}\:\mathrm{any}\:\mathrm{unit}\:\mathrm{vector}\:\mathrm{they}\:\mathrm{are}\:\mathrm{millions}\:\mathrm{of}\:\mathrm{them}\:\mathrm{so} \\ $$$$\mathrm{i}\:\mathrm{guess}\:\mathrm{we}\:\mathrm{can}\:\mathrm{not}\:\mathrm{get}\:\mathrm{a}\:\mathrm{specific}\:\boldsymbol{\mathrm{A}}×\:\boldsymbol{\mathrm{B}}\:\mathrm{vector}\:\mathrm{but}\:\mathrm{we}\:\mathrm{can}\:\mathrm{get}\:\mid\boldsymbol{\mathrm{A}}×\boldsymbol{\mathrm{B}}\mid\:\mathrm{thus}: \\ $$$$\:\mid\boldsymbol{\mathrm{A}}×\boldsymbol{\mathrm{B}}\mid\:=\:\mid\boldsymbol{\mathrm{A}}\mid\mid\boldsymbol{\mathrm{B}}\mid\:\mathrm{sin}\:\theta\:\mid\:\overset{\wedge} {\boldsymbol{\mathrm{n}}}\mid…
Question Number 126398 by sahnaz last updated on 20/Dec/20 $$\overset{\rightarrow} {\mathrm{r}}\left(\mathrm{t}\right)=\mathrm{4sin}^{\mathrm{2}} \mathrm{t}\overset{\rightarrow} {\mathrm{i}}+\mathrm{4cos}^{\mathrm{2}} \mathrm{t}\overset{\rightarrow} {\mathrm{j}}−\mathrm{3}\overset{\rightarrow} {\mathrm{k}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 60628 by cesar.marval.larez@gmail.com last updated on 23/May/19 $${Who}\:\:{know}\:{dynamics}\:{about}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 60576 by hovea cw last updated on 22/May/19 $$\mathrm{two}\:\mathrm{faire}\:\mathrm{dices}\:\mathrm{are}\:\mathrm{tossed}\:\mathrm{together} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\: \\ $$$$\mathrm{total}\:\mathrm{score}\:\mathrm{is}\:\mathrm{atmost}\:\mathrm{4} \\ $$$$ \\ $$ Commented by hovea cw last updated…