Question Number 209944 by Tawa11 last updated on 26/Jul/24 Commented by mr W last updated on 26/Jul/24 $${f}=\frac{\mathrm{100}×\mathrm{6}}{\mathrm{2}\pi×\mathrm{2}.\mathrm{5}}=\mathrm{38}.\mathrm{2}\:{Hz} \\ $$ Terms of Service Privacy Policy…
Question Number 209886 by Tawa11 last updated on 24/Jul/24 Commented by Tawa11 last updated on 25/Jul/24 $$\mathrm{g}\:=\:\mathrm{9}.\mathrm{8}\:\mathrm{ms}^{−\mathrm{2}} \\ $$ Answered by mr W last updated…
Question Number 209847 by SonGoku last updated on 23/Jul/24 Answered by mr W last updated on 23/Jul/24 $${depth}\:={h} \\ $$$${h}=\sqrt{\mathrm{23}^{\mathrm{2}} −\left(\frac{\mathrm{53}−\mathrm{41}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{493}}=\mathrm{22}.\mathrm{20}{m} \\ $$$${cross}\:{section}\:{area}\:={A} \\…
Question Number 209856 by SonGoku last updated on 23/Jul/24 Commented by SonGoku last updated on 23/Jul/24 $$\mathrm{How}\:\mathrm{do}\:\mathrm{I}\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{this}\:\mathrm{figure}? \\ $$ Commented by mr W last updated…
Question Number 209631 by York12 last updated on 17/Jul/24 $$\mathrm{Let}\:{u}_{{n}} \:\mathrm{be}\:\mathrm{a}\:\mathrm{set}\:\mathrm{satisfying}\:{u}_{\mathrm{1}} =\mathrm{1}\:\&\:{u}_{{n}+\mathrm{1}} ={u}_{{n}} +\frac{\mathrm{ln}\:{n}}{{u}_{{n}} }\:\:,\:\forall\:{n}\:\geqslant\mathrm{1} \\ $$$$\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that}\:{u}_{\mathrm{2023}} >\sqrt{\mathrm{2023}.\mathrm{ln}\:\mathrm{2023}}. \\ $$$$\mathrm{2}.\:\mathrm{Find}:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{u}_{{n}} .\mathrm{ln}\:{n}}{{n}}. \\ $$ Terms…
Question Number 209432 by Tawa11 last updated on 10/Jul/24 A body is projected vertically upwards with a speed of 20m/s. Find the time in seconds…
Question Number 209430 by Tawa11 last updated on 09/Jul/24 Answered by mr W last updated on 10/Jul/24 $$\left.\mathrm{1}\right) \\ $$$$\omega_{\mathrm{0}} =\sqrt{\frac{{k}}{{m}}} \\ $$$$\zeta=\frac{{c}}{\mathrm{2}\sqrt{{mk}}} \\ $$$$\omega_{{r}}…
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Question Number 209193 by Tawa11 last updated on 03/Jul/24 A pin 6cm high is placed in front of a diverging lens of focal length 15cm,…
Question Number 209129 by Adeyemi889 last updated on 02/Jul/24 Answered by A5T last updated on 02/Jul/24 $$\frac{{x}^{\mathrm{3}} +\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}={x}+\frac{{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\frac{{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{{A}}{{x}−\mathrm{1}}+\frac{{B}}{{x}+\mathrm{1}}\Rightarrow{x}+\mathrm{3}=\left({A}+{B}\right){x}+{A}−{B}…