Question Number 172604 by Mikenice last updated on 29/Jun/22 Answered by mr W last updated on 29/Jun/22 $${v}_{\mathrm{1}} =\sqrt{\mathrm{2}{gh}_{\mathrm{1}} }=\sqrt{\mathrm{2}×\mathrm{9}.\mathrm{8}×\mathrm{4}.\mathrm{9}}=\mathrm{9}.\mathrm{8}\:{m}/{s} \\ $$$${t}_{\mathrm{1}} =\frac{{v}_{\mathrm{1}} }{{g}}=\frac{\mathrm{9}.\mathrm{8}}{\mathrm{9}.\mathrm{8}}=\mathrm{1}\:{s} \\…
Question Number 172603 by Mikenice last updated on 29/Jun/22 Answered by mr W last updated on 29/Jun/22 Commented by mr W last updated on 29/Jun/22…
Question Number 106997 by Dwaipayan Shikari last updated on 08/Aug/20 Commented by mr W last updated on 08/Aug/20 $${mv}−{m}×\frac{−{v}}{\mathrm{2}}=\int_{\mathrm{0}} ^{{T}} {fdt}={FT} \\ $$$$\Rightarrow{F}=\frac{\mathrm{3}{mv}}{\mathrm{2}{T}} \\ $$…
Question Number 106928 by aurpeyz last updated on 07/Aug/20 $${Imagine}\:{a}\:{planet}\:{having}\:{a}\:{mass}\:{twice}\:{that} \\ $$$${of}\:{the}\:{earth}\:{and}\:{a}\:{radius}\:{equal}\:{to}\:\mathrm{1}.\mathrm{414} \\ $$$${times}\:{that}\:{of}\:{the}\:{earth}.\:{Determine}\:{the} \\ $$$${acceleration}\:{due}\:{to}\:{gravity}\:{at}\:{its}\:{surface}. \\ $$ Answered by JDamian last updated on 07/Aug/20…
Question Number 106921 by aurpeyz last updated on 07/Aug/20 Commented by aurpeyz last updated on 07/Aug/20 $${i}\:{didnt}\:{get}\:{the}\:{answer}\:{to}\:{the}\:{last}\:{part}.\:{pls} \\ $$$${help} \\ $$ Answered by mr W…
Question Number 106919 by aurpeyz last updated on 07/Aug/20 $$\mathrm{if}\:\mathrm{the}\:\mathrm{wheel}\:\mathrm{of}\:\mathrm{a}\:\mathrm{car}\:\mathrm{moved}\:\mathrm{56rev}.\: \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{the}\:\mathrm{car}\:\mathrm{moved}\:\mathrm{in}\:\mathrm{14s}? \\ $$ Commented by Dwaipayan Shikari last updated on 07/Aug/20 $$\mathrm{56}{rev}/{s}?? \\ $$…
Question Number 41255 by ajfour last updated on 04/Aug/18 Commented by ajfour last updated on 04/Aug/18 $${The}\:{natural}\:{length}\:{of}\:{spring}\:{is} \\ $$$${l}\:{and}\:{if}\:{realeased}\:{as}\:{shown},\:{find} \\ $$$${speed}\:{of}\:{ring}\:{as}\:{it}\:{reaches}\:{the} \\ $$$${ground}.\:\mu\:{is}\:{the}\:{friction}\:{coefficient} \\ $$$${between}\:{vertical}\:{rod}\:{and}\:{the}\:{ring}.…
Question Number 41233 by ajfour last updated on 03/Aug/18 Commented by ajfour last updated on 03/Aug/18 $${Find}\:\boldsymbol{\mathrm{v}}\:{as}\:{a}\:{function}\:{of}\:\theta. \\ $$$${Initially}\:\theta=\mathrm{0}°\:. \\ $$ Answered by MrW3 last…
Question Number 40939 by ajfour last updated on 29/Jul/18 Commented by ajfour last updated on 29/Jul/18 $${Related}\:{to}\:{Q}.\mathrm{40920} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 40928 by rahul 19 last updated on 29/Jul/18 Answered by MrW3 last updated on 29/Jul/18 $${length}\:{of}\:{rope}\:{l}=\mathrm{2}\pi{r}=\pi{R} \\ $$$${let}\:\lambda=\frac{{W}}{{l}}=\frac{{W}}{\pi{R}} \\ $$$$\mathrm{cos}\:\theta=\frac{{r}}{{R}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\theta=\frac{\pi}{\mathrm{3}}=\mathrm{60}° \\…