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Show-that-a-n-r-2-show-clearly-how-you-arrive-at-your-result-

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Question Number 82358 by TawaTawa last updated on 20/Feb/20 $$\mathrm{Show}\:\mathrm{that}:\:\:\:\:\:\:\:\mathrm{a}_{\mathrm{n}} \:\:=\:\:−\:\mathrm{r}\omega^{\mathrm{2}} \:,\:\:\:\mathrm{show}\:\mathrm{clearly}\:\mathrm{how}\:\mathrm{you}\:\mathrm{arrive} \\ $$$$\mathrm{at}\:\mathrm{your}\:\mathrm{result}. \\ $$ Commented by TawaTawa last updated on 20/Feb/20 $$\mathrm{Please}\:\mathrm{help} \\…

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Question-147778

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Question Number 147778 by mnjuly1970 last updated on 23/Jul/21 Answered by liberty last updated on 23/Jul/21 $$\left(\bullet\right)\angle{ABC}=\angle{CBD}=\theta \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{3}}{{AB}}\: \\ $$$$\left(\bullet\right)\mathrm{cos}\:\mathrm{2}\theta=\frac{\mathrm{7}}{{AB}}=\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \theta \\ $$$$\Rightarrow\frac{\mathrm{7}}{{AB}}=\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{9}}{{AB}^{\mathrm{2}} }\right)\:\Rightarrow{AB}={x}…

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The-horizontal-range-of-a-projectile-is-R-and-the-maximum-height-attained-by-it-is-H-A-strong-wind-now-begins-to-blow-in-the-direction-of-horizontal-motion-of-projectile-giving-it-a-constant-horizon

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Question Number 16691 by Tinkutara last updated on 25/Jun/17 $$\mathrm{The}\:\mathrm{horizontal}\:\mathrm{range}\:\mathrm{of}\:\mathrm{a}\:\mathrm{projectile} \\ $$$$\mathrm{is}\:{R}\:\mathrm{and}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height}\:\mathrm{attained} \\ $$$$\mathrm{by}\:\mathrm{it}\:\mathrm{is}\:{H}.\:\mathrm{A}\:\mathrm{strong}\:\mathrm{wind}\:\mathrm{now}\:\mathrm{begins}\:\mathrm{to} \\ $$$$\mathrm{blow}\:\mathrm{in}\:\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{horizontal} \\ $$$$\mathrm{motion}\:\mathrm{of}\:\mathrm{projectile},\:\mathrm{giving}\:\mathrm{it}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\mathrm{horizontal}\:\mathrm{acceleration}\:\mathrm{equal}\:\mathrm{to}\:{g}. \\ $$$$\mathrm{Under}\:\mathrm{the}\:\mathrm{same}\:\mathrm{conditions}\:\mathrm{of}\:\mathrm{projection}, \\ $$$$\mathrm{the}\:\mathrm{new}\:\mathrm{range}\:\mathrm{will}\:\mathrm{be} \\…

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Question-16682

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Question Number 16682 by tawa tawa last updated on 25/Jun/17 Answered by sandy_suhendra last updated on 25/Jun/17 $$\mathrm{F}=\mathrm{0}.\mathrm{2}\:\mathrm{N}\: \\ $$$$\Delta\mathrm{x}=\mathrm{1}\:\mathrm{mm}\:=\:\mathrm{10}^{−\mathrm{3}} \mathrm{m} \\ $$$$\mathrm{k}\:=\:\frac{\mathrm{F}}{\Delta\mathrm{x}}\:=\:\frac{\mathrm{0}.\mathrm{2}}{\mathrm{10}^{−\mathrm{3}} }\:=\:\mathrm{200}\:\mathrm{N}/\mathrm{m} \\…

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