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The-velocity-of-a-particle-moving-in-straight-line-is-given-by-the-graph-shown-here-Draw-the-acceleration-position-graph-

Question Number 13745 by Tinkutara last updated on 23/May/17 $$\mathrm{The}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{in} \\ $$$$\mathrm{straight}\:\mathrm{line}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{shown} \\ $$$$\mathrm{here}.\:\mathrm{Draw}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{position} \\ $$$$\mathrm{graph}. \\ $$ Commented by Tinkutara last updated on 23/May/17…

A-light-wire-which-obeys-hooke-s-law-hangs-vertically-on-a-fixed-support-The-wire-has-an-unstretched-lenght-of-15cm-The-lenght-of-the-wire-however-increase-to-20cm-when-a-load-of-0-5kg-is-attached-

Question Number 13727 by tawa tawa last updated on 22/May/17 $$\mathrm{A}\:\mathrm{light}\:\mathrm{wire}\:\mathrm{which}\:\mathrm{obeys}\:\mathrm{hooke}'\mathrm{s}\:\mathrm{law}\:\mathrm{hangs}\:\mathrm{vertically}\:\mathrm{on}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{support}. \\ $$$$\mathrm{The}\:\mathrm{wire}\:\mathrm{has}\:\mathrm{an}\:\mathrm{unstretched}\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{15cm}.\:\:\mathrm{The}\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wire}\:\mathrm{however} \\ $$$$\mathrm{increase}\:\mathrm{to}\:\mathrm{20cm}\:\mathrm{when}\:\mathrm{a}\:\mathrm{load}\:\mathrm{of}\:\mathrm{0}.\mathrm{5kg}\:\mathrm{is}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{it}\:\mathrm{lower}\:\mathrm{end}\:.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{wire}\:\mathrm{when}\:\mathrm{the}\:\mathrm{load}\:\mathrm{is}\:\mathrm{at}\:\mathrm{rest}\:?.\: \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{load}\:\mathrm{is}\:\mathrm{pulled}\:\mathrm{down}\:\mathrm{until}\:\mathrm{the}\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wire}\:\mathrm{is}\:\mathrm{22cm}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{new} \\ $$$$\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{wire}\:\left(\mathrm{g}\:=\:\mathrm{9}.\mathrm{8}\:\mathrm{m}/\mathrm{s}\right). \\ $$ Commented by…

Question-13688

Question Number 13688 by aishadenge last updated on 22/May/17 Answered by ajfour last updated on 22/May/17 $$\frac{\partial}{\partial{R}_{\mathrm{1}} }\left(\frac{\mathrm{1}}{{R}}\right)=−\frac{\mathrm{1}}{{R}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:−\frac{\mathrm{1}}{{R}^{\mathrm{2}} }\left(\frac{\partial{R}}{\partial{R}_{\mathrm{1}} }\right)=−\frac{\mathrm{1}}{{R}_{\mathrm{1}} ^{\mathrm{2}}…

Question-13687

Question Number 13687 by christine last updated on 22/May/17 Commented by mrW1 last updated on 22/May/17 $${The}\:{question}\:{is}\:{wrong}.\:{It}\:{should}\:{be} \\ $$$$\boldsymbol{{z}}=\boldsymbol{{e}}^{−\boldsymbol{{x}}} \left(\boldsymbol{{y}}\mathrm{cos}\:\boldsymbol{{x}}+\boldsymbol{{x}}\mathrm{sin}\:\boldsymbol{{y}}\right) \\ $$$$ \\ $$$${see}\:{Q}\mathrm{13544}. \\…

Question-79177

Question Number 79177 by TawaTawa last updated on 23/Jan/20 Answered by mr W last updated on 23/Jan/20 $${M}\frac{{L}}{{T}^{\mathrm{2}} }=\left[{b}\right]\frac{{L}}{{T}}=\frac{{L}^{\mathrm{2}} +\left[{d}\right]^{\mathrm{2}} }{\left[{c}\right]} \\ $$$$\Rightarrow\left[{b}\right]=\frac{{M}}{{T}},\:{e}.{g}.\:\left({kg}/{s}\right) \\ $$$$\Rightarrow\left[{d}\right]={L},\:{e}.{g}.\:\left({m}\right)…