Question Number 103792 by aurpeyz last updated on 17/Jul/20 Answered by Dwaipayan Shikari last updated on 17/Jul/20 $${v}_{{f}} ^{\mathrm{2}} ={v}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}{as} \\ $$$$\mathrm{400}=−\mathrm{2}{a}.\mathrm{120} \\…
Question Number 169317 by BHOOPENDRA last updated on 28/Apr/22 Commented by BHOOPENDRA last updated on 28/Apr/22 $${find}\:{time}\:{of}\:{flight}\:{also}? \\ $$ Commented by mr W last updated…
Question Number 103739 by aurpeyz last updated on 17/Jul/20 Answered by Rio Michael last updated on 17/Jul/20 $$\:\mathrm{at}\:\mathrm{limiting}\:\mathrm{equilibrium},\:{F}_{{f}} \:\:+\:\mathrm{mg}\:\mathrm{sin}\:\theta\:=\:{P}\: \\ $$$$\mathrm{but}\:{F}_{{f}} \:=\:{N}\mu\:\:\mathrm{but}\:{N}\:=\:\mathrm{mg}\:\mathrm{cos}\:\theta\:=\:\left(\mathrm{163}.\mathrm{5}\right)\mathrm{cos}\:\mathrm{14}.\mathrm{5}\:=\:\mathrm{158}.\mathrm{29}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\: \\ $$$$\mathrm{100}^{\mathrm{th}} \:=\:\mathrm{200}\:{N}…
Question Number 169205 by novayanti_sinaga93 last updated on 26/Apr/22 $$\sqrt{\mathrm{91}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 38094 by ajfour last updated on 21/Jun/18 Commented by ajfour last updated on 21/Jun/18 $${A}\:{rod}\:{hangs}\:{from}\:{ceiling}\:{by}\:{two} \\ $$$${two}\:{strings}\:{of}\:{length}\:\boldsymbol{{a}},\:\boldsymbol{{b}}\:{from} \\ $$$${points}\:{on}\:{ceiling}\:{separated}\:{by} \\ $$$${distance}\:\boldsymbol{{d}}.\:{The}\:{rod}\:{has}\:{length}\:\boldsymbol{{l}}. \\ $$$$\boldsymbol{{F}}{ind}\:\boldsymbol{\theta}.…
Question Number 103586 by aurpeyz last updated on 15/Jul/20 Answered by OlafThorendsen last updated on 16/Jul/20 $$\overset{\rightarrow} {\mathrm{F}}_{\mathrm{1}} +\overset{\rightarrow} {\mathrm{F}}_{\mathrm{2}} +\overset{\rightarrow} {\mathrm{F}}_{\mathrm{3}} +\overset{\rightarrow} {\mathrm{P}}\:=\:\overset{\rightarrow} {\mathrm{0}}…
Question Number 103584 by aurpeyz last updated on 15/Jul/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 103585 by aurpeyz last updated on 15/Jul/20 Answered by Worm_Tail last updated on 16/Jul/20 $${CW}={ACW} \\ $$$$\:\:\:\left(\mathrm{430}×\mathrm{9}.\mathrm{8}×\mathrm{2}.\mathrm{5}\right)+\left(\mathrm{85}×\mathrm{9}.\mathrm{8}×\frac{\sqrt{\mathrm{1}.\mathrm{67}^{\mathrm{2}} +\mathrm{2}.\mathrm{5}^{\mathrm{2}} }}{\mathrm{2}}{sin}\left({tan}^{−\mathrm{1}} \frac{\mathrm{2}.\mathrm{5}}{\mathrm{1}.\mathrm{67}}\right)\right)={T}\left(\mathrm{1}.\mathrm{67}\right) \\ $$$$\:\:\:\left(\mathrm{10535}\right)+\left(\mathrm{1041}.\mathrm{25}\right)={T}\left(\mathrm{1}.\mathrm{67}\right) \\…
Question Number 103582 by aurpeyz last updated on 15/Jul/20 Commented by mr W last updated on 16/Jul/20 $${no}\:{unique}\:{solution}\:{possible}! \\ $$ Commented by mr W last…
Question Number 103450 by aurpeyz last updated on 15/Jul/20 Answered by Worm_Tail last updated on 15/Jul/20 $$\mu={tan}\theta \\ $$$$\mathrm{0}.\mathrm{35}={tan}\theta \\ $$$$\theta={tan}^{−\mathrm{1}} \mathrm{0}.\mathrm{35}=\mathrm{19}.\mathrm{29}° \\ $$ Commented…