Question Number 53145 by ajfour last updated on 18/Jan/19 Commented by ajfour last updated on 18/Jan/19 $${If}\:{entire}\:{string}\:{length}\:{is}\:{l},\:{find}\:\theta \\ $$$${in}\:{terms}\:{of}\:{l},{a},\:{r},\:{and}\:{R}. \\ $$$${Or}\:{find}\:{l}\:{from}\:{a},{r},{R},\:{and}\:\theta. \\ $$ Commented by…
Question Number 52938 by ajfour last updated on 15/Jan/19 Commented by ajfour last updated on 16/Jan/19 $${Find}\:\boldsymbol{{x}},\:{tension}\:\boldsymbol{{T}},\:{and}\:\boldsymbol{\theta}. \\ $$$${Assume}\:{friction}\:{is}\:{not}\:{present} \\ $$$${and}\:{rod}\:{is}\:{in}\:{equilibrium}.\:\:\:\:\: \\ $$ Commented by…
Question Number 52891 by ajfour last updated on 14/Jan/19 Commented by ajfour last updated on 15/Jan/19 $${Find}\:{tensions}\:{in}\:{the}\:{two}\:{strings}. \\ $$$${let}\:{for}\:{example},\:\:{M}=\mathrm{5}{kg},\:{m}=\:\mathrm{3}{kg} \\ $$$${R}=\:\mathrm{20}{cm}\:,\:{p}\:=\:\mathrm{40}{cm}\:,\:{r}=\:\mathrm{10}{cm}, \\ $$$${q}\:=\:\mathrm{30}{cm}\:\:\left({to}\:{match}\:{our}\:{answers}\right). \\ $$…
Question Number 52847 by peter frank last updated on 14/Jan/19 Commented by peter frank last updated on 14/Jan/19 Commented by peter frank last updated on…
Question Number 52825 by ajfour last updated on 13/Jan/19 Commented by ajfour last updated on 13/Jan/19 $${Find}\:\theta\:.\:\:\:{Assume}\:{no}\:{friction}\:{and} \\ $$$${a}\:{static}\:{situation}. \\ $$ Commented by tanmay.chaudhury50@gmail.com last…
Question Number 52814 by ajfour last updated on 13/Jan/19 Commented by ajfour last updated on 13/Jan/19 $${Initially}\:{rod}\:{whose}\:{one}\:{end}\:{is} \\ $$$${pivoted}\:{to}\:{centre}\:{of}\:{disc},\:{is}\:{kept} \\ $$$${vertical}\:{and}\:{released},\:{Find}\:{v}\left(\theta\right). \\ $$$${Friction}\:{is}\:{sufficient}\:{say},\: \\ $$$${coefficient}\:{is}\:\mu.…
Question Number 52795 by peter frank last updated on 13/Jan/19 Answered by peter frank last updated on 13/Jan/19 Answered by peter frank last updated on…
Question Number 52790 by peter frank last updated on 13/Jan/19 Answered by peter frank last updated on 13/Jan/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 52787 by peter frank last updated on 13/Jan/19 Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jan/19 $${flow}\:{of}\:{fluid}\:{in}\:{pipe}\:{formula} \\ $$$$\frac{{dV}}{{dt}}=\frac{\bigtriangleup{p}×\pi{r}^{\mathrm{4}} }{\mathrm{8}\eta{l}}=\frac{\bigtriangleup{p}}{\frac{\mathrm{8}\eta{l}}{\pi{r}^{\mathrm{4}} }} \\ $$$${compare}\:{it}\:{with}\:{flow}\:{of}\:{current}\:{in}\:{resistance}.. \\…
Question Number 52786 by peter frank last updated on 13/Jan/19 Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jan/19 $${mw}^{\mathrm{2}} {r}=\frac{{GMm}}{{r}^{\mathrm{2}} }\:\left[\:{M}={mass}\:{of}\:{sun},\:\:{m}={mass}\:{of}\:{planet}\right] \\ $$$$\left(\frac{\mathrm{2}\pi}{{T}}\right)^{\mathrm{2}} =\frac{{GM}}{{r}^{\mathrm{3}} }\:\left[{r}={distance}\:{between}\:{planet}\:{and}\:{sun}\right] \\…