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Category: Mechanics

A-cubical-block-of-ice-of-mass-m-and-edge-L-is-placed-in-a-large-tray-of-mass-M-If-the-ice-block-melts-how-far-does-the-centre-of-mass-of-the-system-ice-tray-come-down-a-ml-m-M-b-2ml-

Question Number 61205 by necx1 last updated on 30/May/19 $${A}\:{cubical}\:{block}\:{of}\:{ice}\:{of}\:{mass}\:{m}\:{and} \\ $$$${edge}\:{L}\:{is}\:{placed}\:{in}\:{a}\:{large}\:{tray}\:{of}\:{mass} \\ $$$${M}.{If}\:{the}\:{ice}\:{block}\:{melts},{how}\:{far}\:{does} \\ $$$${the}\:{centre}\:{of}\:{mass}\:{of}\:{the}\:{system}\:“{ice}\:+\:{tray}'' \\ $$$${come}\:{down}\:? \\ $$$$ \\ $$$$\left.{a}\left.\right)\left.\frac{{ml}}{{m}+{M}}\left.\:\:{b}\right)\frac{\mathrm{2}{ml}}{{m}+{M}}\:\:{c}\right)\frac{{ml}}{\mathrm{2}\left({m}+{M}\right)}\:\:{d}\right){none} \\ $$ Commented…

If-there-is-a-positive-error-in-the-measurement-of-velocity-of-a-body-then-the-error-in-the-measure-ment-of-kinetic-energy-is-

Question Number 126603 by Ar Brandon last updated on 22/Dec/20 $$\mathrm{If}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{error}\:\mathrm{in}\:\mathrm{the}\:\mathrm{measurement} \\ $$$$\mathrm{of}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{body},\:\mathrm{then}\:\mathrm{the}\:\mathrm{error}\:\mathrm{in}\:\mathrm{the}\:\mathrm{measure}- \\ $$$$\mathrm{ment}\:\mathrm{of}\:\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{is} \\ $$ Answered by Olaf last updated on 22/Dec/20 $$\mathrm{E}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}}…

A-particle-of-mass-m-moves-under-the-central-repulsive-force-mb-r-3-and-is-initially-moving-at-a-distance-a-from-the-origin-of-a-force-with-velocity-v-at-right-angle-to-a-show-that-

Question Number 191868 by Spillover last updated on 02/May/23 $${A}\:{particle}\:{of}\:{mass}\:{m}\:{moves}\:{under}\:{the}\:{central} \\ $$$${repulsive}\:{force}\:\frac{{mb}}{{r}^{\mathrm{3}} }\:\:{and}\:{is}\:{initially}\:{moving} \\ $$$${at}\:{a}\:{distance}\:'{a}'\:\:{from}\:{the}\:{origin}\:{of}\:\:{a}\:{force} \\ $$$${with}\:{velocity}\:\:'{v}'\:{at}\:{right}\:{angle}\:{to}\:\:'{a}'. \\ $$$${show}\:{that}\:\:\: \\ $$$$\:\:\:\:\:{r}\mathrm{cos}\:{p}\theta={a}\:\:{where}\:{p}\:=\frac{{b}}{{a}^{\mathrm{2}} {v}^{\mathrm{2}} }+\mathrm{1}. \\ $$$$…