Question Number 103421 by aurpeyz last updated on 15/Jul/20 Answered by Worm_Tail last updated on 15/Jul/20 $${v}_{{y}} ^{\mathrm{2}} =\left({usin}\theta\right)^{\mathrm{2}} −\mathrm{2}{gs}_{{y}} \Rightarrow\mathrm{3}.\mathrm{7}^{\mathrm{2}} =\left({usin}\theta\right)^{\mathrm{2}} −\left(\mathrm{2}×\mathrm{10}×\mathrm{7}.\mathrm{5}\right) \\ $$$$\mathrm{13}.\mathrm{69}+\mathrm{150}=\left({usin}\theta\right)^{\mathrm{2}}…
Question Number 103419 by aurpeyz last updated on 15/Jul/20 Answered by Worm_Tail last updated on 15/Jul/20 $${v}_{{y}} ={usin}\theta−{gt}\Rightarrow\mathrm{4}.\mathrm{8}={usin}\theta−\left(\mathrm{10}×\mathrm{0}.\mathrm{8}\right) \\ $$$$\mathrm{4}.\mathrm{8}+\mathrm{8}={usin}\theta\Rightarrow{usin}\theta=\mathrm{12}.\mathrm{8} \\ $$$${h}_{{max}} =\frac{\left({usin}\theta\right)^{\mathrm{2}} }{\mathrm{2}{g}}=\frac{\left(\mathrm{12}.\mathrm{8}\right)^{\mathrm{2}} }{\mathrm{20}}=\mathrm{8}.\mathrm{192}{m}…
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Question Number 103143 by ajfour last updated on 13/Jul/20 Commented by ajfour last updated on 13/Jul/20 $${Q}.\mathrm{103072}\:\:\:\:\left({A}\:{revisit}\right) \\ $$ Answered by ajfour last updated on…
Question Number 37579 by ajfour last updated on 15/Jun/18 $${a}=\frac{−\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:\:{with}\:{x}=\mathrm{1}\:{and}\:{v}=\mathrm{0}\:{at}\:{t}=\mathrm{0} \\ $$$${find}\:{time}\:{that}\:{particle}\:{takes}\:{to} \\ $$$${reach}\:{x}=\mathrm{0}.\mathrm{25}{m}\:. \\ $$ Answered by ajfour last updated on 15/Jun/18 $${since}\:{a}\:<\:\mathrm{0}\:{and}\:{v}_{\mathrm{0}}…
Question Number 103072 by mr W last updated on 12/Jul/20 Commented by ajfour last updated on 12/Jul/20 Commented by mr W last updated on 12/Jul/20…
Question Number 102978 by ajfour last updated on 12/Jul/20 Commented by ajfour last updated on 12/Jul/20 $${The}\:{inclined}\:{plane}\:{of}\:{mass}\:{m}, \\ $$$${length}\:{L}\:{and}\:{the}\:{block}\:{of}\:{mass}\:{m}, \\ $$$${both}\:{are}\:{on}\:{the}\:{verge}\:{of}\:{slipping}. \\ $$$${Find}\:{x}\:{and}\:\mu. \\ $$…
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Question Number 102880 by Dwaipayan Shikari last updated on 11/Jul/20 Commented by Dwaipayan Shikari last updated on 11/Jul/20 $${Sorry}\:{sir}\:.\:{I}\:{did}\:{not}\:{include}\:{it}\:.\:{The}\:{weight}\:{recorder}\:{is}\:{mass} \\ $$$${less}.{The}\:{man}\:{stands}\:{on}\:{it}.{if}\:{Recorder}\:{records}\:\mathrm{72}{kg}\:{as}\:{the} \\ $$$${weight}\:{of}\:{the}\:{man}\:{the}.\:{Here}\:{the}\:{weight}\:{recorder}\:{is}\:{a}\:{trianguler} \\ $$$${platform}.\:{And}\:{it}\:{slides}.\:{Friction}\:{is}\:{along}\:{the}\:{slope}…
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