Question Number 93568 by kikomssn@gmail.com last updated on 13/May/20 Commented by prakash jain last updated on 14/May/20 $$\mathrm{English}\:\mathrm{please} \\ $$ Terms of Service Privacy Policy…
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Question Number 28027 by ajfour last updated on 18/Jan/18 Answered by mrW2 last updated on 19/Jan/18 $${radius}\:{of}\:{base}\:{circle}\:={R} \\ $$$${radius}\:{of}\:{disc}\:{in}\:{distance}\:{x}\:{from}\:{the} \\ $$$${base}\:{is}\:{r}=\frac{{h}−{x}}{{h}}×{R}=\left(\mathrm{1}−\frac{{x}}{{h}}\right){R} \\ $$$${mass}\:{of}\:{disc}\:{is}\:{dm}=\rho\pi{r}^{\mathrm{2}} {dx} \\…
Question Number 159085 by ajfour last updated on 12/Nov/21 Answered by mr W last updated on 15/Nov/21 Commented by mr W last updated on 15/Nov/21…
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Question Number 27899 by Tinkutara last updated on 16/Jan/18 Answered by mrW2 last updated on 17/Jan/18 $${let}\:{e}={max}.\:{elongation}\:{of}\:{spring}\:{after}\:{release} \\ $$$${ke}=\mu{m}_{{A}} {g}\:\:\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{kd}^{\mathrm{2}} −\mu{m}_{{B}} {g}\left({d}+{e}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ke}^{\mathrm{2}} \:\:\:\:…\left({ii}\right)…
Question Number 158874 by ajfour last updated on 09/Nov/21 Commented by ajfour last updated on 09/Nov/21 $${Length}\:{of}\:{rope}\:{is}\:{L},\:{mass}\:{m}. \\ $$$${The}\:{pulley}\:{even}\:{has}\:{mass}\:{m} \\ $$$${and}\:{radius}\:{is}\:{r}.\:{The}\:{bead}\:{has} \\ $$$${mass}\:{m}.\:{After}\:{release}\:{at}\:{t}=\mathrm{0}, \\ $$$${find}\:{v}\left({x}\right);\:{v}\:{being}\:{the}\:{speed}\:{of}…
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Question Number 27677 by Tinkutara last updated on 12/Jan/18 Commented by Tinkutara last updated on 12/Jan/18 Commented by ajfour last updated on 14/Jan/18 Commented by…
Question Number 27564 by Tinkutara last updated on 09/Jan/18 Commented by Tinkutara last updated on 09/Jan/18 Commented by mrW2 last updated on 12/Jan/18 Commented by…