Question Number 54770 by Tawa1 last updated on 10/Feb/19 $$\mathrm{A}\:\mathrm{glass}\:\mathrm{bottle}\:\mathrm{full}\:\mathrm{of}\:\mathrm{mercury}\:\mathrm{has}\:\mathrm{mass}\:\mathrm{500g}.\:\mathrm{On}\:\mathrm{being}\:\mathrm{heated}\:\mathrm{through} \\ $$$$\mathrm{35}°\mathrm{C},\:\:\mathrm{2}.\mathrm{43g}\:\mathrm{of}\:\mathrm{mercury}\:\mathrm{are}\:\mathrm{expelled}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{mercury} \\ $$$$\mathrm{remaining}\:\mathrm{in}\:\mathrm{the}\:\mathrm{bottle}.\:\left(\mathrm{Cubic}\:\mathrm{expansivity}\:\mathrm{of}\:\mathrm{mercury}\:\mathrm{is}\:\:\mathrm{1}.\mathrm{8}\:×\:\mathrm{10}^{−\mathrm{4}} \:\mathrm{K}^{−\mathrm{1}} \:,\right. \\ $$$$\mathrm{linear}\:\mathrm{expansivity}\:\mathrm{of}\:\mathrm{glass}\:\mathrm{is}\:\:\mathrm{8}.\mathrm{0}\:×\:\mathrm{10}^{−\mathrm{6}} \mathrm{K}^{−\mathrm{1}} \:. \\ $$ Terms of Service…
Question Number 54767 by pieroo last updated on 10/Feb/19 $$\mathrm{A}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upwards} \\ $$$$\mathrm{from}\:\mathrm{a}\:\mathrm{cliff}\:\mathrm{20m}\:\mathrm{high}.\:\mathrm{After}\:\mathrm{a}\:\mathrm{time}\:\mathrm{of}\:\mathrm{3}\:\mathrm{s} \\ $$$$\mathrm{it}\:\mathrm{passes}\:\mathrm{the}\:\mathrm{edge}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cliff}\:\mathrm{on}\:\mathrm{its}\:\mathrm{way} \\ $$$$\mathrm{down}.\:\mathrm{Calculate} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{projection} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{the}\:\mathrm{speed}\:\mathrm{when}\:\mathrm{it}\:\mathrm{hits}\:\mathrm{the}\:\mathrm{ground} \\ $$$$\left.\mathrm{c}\right)\:\mathrm{the}\:\mathrm{times}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{10m}\:\mathrm{above}\:\mathrm{the} \\ $$$$\mathrm{top}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cliff} \\…
Question Number 54745 by arvinddayama00@gmail.com last updated on 10/Feb/19 $$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} +}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }={a}\left({x}−{y}\right) \\ $$$$\frac{{dy}}{{dx}}=\sqrt{\frac{\mathrm{1}−{y}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${without}\:{solve}\:{put}\:{x}={sin}\theta{and} \\ $$$${y}={sin}\phi \\ $$$$ \\ $$$$ \\…
Question Number 54716 by Tawa1 last updated on 09/Feb/19 $$\mathrm{A}\:\mathrm{light}\:\mathrm{horizontal}\:\mathrm{meter}\:\mathrm{rule}\:\:\mathrm{PQR}\:\:\mathrm{has}\:\mathrm{the}\:\mathrm{end}\:\:\mathrm{P}\:\mathrm{fixed}\:\mathrm{to}\: \\ $$$$\mathrm{vertical}\:\mathrm{wall},\:\mathrm{while}\:\mathrm{a}\:\mathrm{weight}\:\mathrm{of}\:\:\mathrm{5N}\:\mathrm{is}\:\mathrm{suspended}\:\mathrm{from}\:\mathrm{the}\:\mathrm{end} \\ $$$$\mathrm{R}.\:\mathrm{A}\:\mathrm{light}\:\mathrm{string}\:\mathrm{QS}\:\:\mathrm{of}\:\mathrm{length}\:\:\mathrm{50cm}\:\mathrm{fixed}\:\mathrm{to}\:\mathrm{the}\:\mathrm{wall}\:\mathrm{at}\:\mathrm{S}\:\mathrm{is}\: \\ $$$$\mathrm{used}\:\mathrm{to}\:\mathrm{maintain}\:\mathrm{the}\:\mathrm{meter}\:\mathrm{rule}\:\mathrm{in}\:\mathrm{equilibrium}. \\ $$$$\mathrm{If}\:\:\mathrm{PQ}\:=\:\mathrm{40}\:\mathrm{cm},\:\:\:\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{string}\:\mathrm{is}\:? \\ $$ Commented by Tawa1 last updated…
Question Number 185781 by Mingma last updated on 27/Jan/23 Answered by ajfour last updated on 27/Jan/23 $${let}\:{BE}=\mathrm{4}{p}\:,\:\:{BD}=\mathrm{4}{q} \\ $$$${F}\left(\mathrm{3}{p},{q}\right)\:\:;\:\:\:{G}\left(\mathrm{3}{p}+\frac{\mathrm{3}}{\mathrm{4}},\:{q}+\mathrm{1}\right) \\ $$$${FG}=\sqrt{\frac{\mathrm{9}}{\mathrm{16}}+\mathrm{1}}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$ Terms of…
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Question Number 54647 by gunawan last updated on 08/Feb/19 $$\mathrm{show}\:\mathrm{that} \\ $$$${a}.\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\Sigma}}\:{r}^{\mathrm{3}} ._{{n}} {C}_{{r}} ={n}^{\mathrm{2}} \left({n}+\mathrm{3}\right).\mathrm{2}^{{n}−\mathrm{3}} \\ $$$${b}.\:_{{n}} {C}_{\mathrm{0}} ._{{n}} {C}_{\mathrm{1}} +_{{n}} {C}_{\mathrm{1}}…
Question Number 120160 by 77731 last updated on 29/Oct/20 Commented by JDamian last updated on 29/Oct/20 https://m.youtube.com/watch?v=9kfmk7jEuvQ Answered by mindispower last updated on 29/Oct/20 $$\frac{\mathrm{1}}{\mathrm{2}^{{m}}…