Question Number 74678 by necxxx last updated on 28/Nov/19 $${A}\:\mathrm{5}{kg}\:{block}\:{rests}\:{on}\:{a}\:\mathrm{30}°\:{incline}.{The} \\ $$$${coefficient}\:{of}\:{static}\:{friction}\:{between} \\ $$$${the}\:{block}\:{and}\:{incline}\:{is}\:\mathrm{0}.\mathrm{20}.{How}\:{large} \\ $$$${a}\:{horizontal}\:{force}\:{must}\:{push}\:{on}\:{the} \\ $$$${block}\:{if}\:{the}\:{block}\:{is}\:{to}\:{be}\:{on}\:{the}\:{verge}\:{of} \\ $$$${sliding}\:\left({a}\right){up}\:{the}\:{incline}\:\left({b}\right)\:{down}\:{the} \\ $$$${incline}? \\ $$ Answered…
Question Number 74594 by ajfour last updated on 27/Nov/19 Commented by ajfour last updated on 27/Nov/19 $${Reposted}:\:{Q}.\mathrm{74557} \\ $$$${A}\:{particle}\:{of}\:{mass}\:{m}\:{tied}\:{to}\:{O} \\ $$$${with}\:{a}\:{string}\:{of}\:{length}\:{b}\:{and} \\ $$$${released}\:{at}\:{t}=\mathrm{0}\:{at}\:{the}\:{rim}\:{of} \\ $$$${a}\:{hollow}\:{hemisphere}\:{crucible}.…
Question Number 74560 by ajfour last updated on 26/Nov/19 Commented by ajfour last updated on 26/Nov/19 $${Let}\:{height}\:{be}\:\boldsymbol{{b}}.\:{Relate}\:\Omega\:{to}\:\:\frac{{d}\omega}{{dt}}\:. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 74557 by ajfour last updated on 26/Nov/19 Commented by ajfour last updated on 26/Nov/19 $${In}\:{a}\:{fixed}\:{frictionless}\:{crucible} \\ $$$${in}\:{the}\:{form}\:{of}\:{hollow}\:{hemisphere}, \\ $$$${radius}\:{R},\:{a}\:{small}\:{mass}\:{m}\:{is}\: \\ $$$$\:{released}\:{with}\:{a}\:{string}\:\left({length}\right. \\ $$$$\left.\:{b}\:<\:\sqrt{\mathrm{2}}\:{R}\:\right)\:{attaching}\:{it}\:{to}\:{a}\:{point}…
Question Number 74522 by necxxx last updated on 25/Nov/19 $${A}\:{rope}\:{inclined}\:{at}\:{angle}\:\mathrm{37}°\:{to}\:{the}\: \\ $$$${horizontal}\:{is}\:{used}\:{to}\:{drag}\:{a}\:\mathrm{50}{kg}\:{block} \\ $$$${along}\:{a}\:{level}\:{floor}\:{with}\:{an}\:{acceleration} \\ $$$${of}\:\mathrm{1}{m}/{s}^{\mathrm{2}} \:.{The}\:{coefficient}\:{of}\:{friction} \\ $$$${between}\:{the}\:{block}\:{and}\:{the}\:{floor}\:{is}\:\mathrm{0}.\mathrm{2}. \\ $$$${What}\:{is}\:{the}\:{tension}\:{in}\:{the}\:{rope}? \\ $$ Commented by…
Question Number 74419 by peter frank last updated on 24/Nov/19 Commented by peter frank last updated on 29/Nov/19 $${please} \\ $$ Terms of Service Privacy…
Question Number 74418 by peter frank last updated on 24/Nov/19 Answered by ajfour last updated on 24/Nov/19 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{\:^{\overset{{N}} {\nwarrow}} \underset{\underset{{mg}} {\downarrow}} {\mathbb{C}}\rightarrow_{{mv}^{\mathrm{2}} /{R}} } {\diagup}…
Question Number 74415 by Learner-123 last updated on 24/Nov/19 Commented by Learner-123 last updated on 24/Nov/19 $${please}\:{solve}\:{this}\:{by}\:{conservation}\:{of} \\ $$$${energy}. \\ $$ Answered by ajfour last…
Question Number 74308 by Learner-123 last updated on 21/Nov/19 Answered by Tanmay chaudhury last updated on 22/Nov/19 $${P}×\mathrm{2}{R}={I}\alpha \\ $$$$\alpha=\frac{{w}−{w}_{\mathrm{0}} }{{t}} \\ $$$$\frac{{w}−{w}_{\mathrm{0}} }{{t}}=\frac{{P}×\mathrm{2}{R}}{{I}} \\…
Question Number 74274 by Learner-123 last updated on 21/Nov/19 Answered by mr W last updated on 21/Nov/19 $${ring}: \\ $$$${r}=\sqrt{{l}^{\mathrm{2}} −{h}^{\mathrm{2}} } \\ $$$${M}=\mathrm{2}\pi{r}\rho \\…