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z-z-2-3x-3y-0-show-that-2-z-x-2-2-z-y-2-2z-x-1-z-2-x-3-

Question Number 89236 by 675480065 last updated on 16/Apr/20 $$\mathrm{z}\left(\mathrm{z}^{\mathrm{2}} +\mathrm{3x}\right)+\mathrm{3y}=\mathrm{0} \\ $$$$\mathrm{show}\:\mathrm{that} \\ $$$$\frac{\partial^{\mathrm{2}} \mathrm{z}}{\partial\mathrm{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} \mathrm{z}}{\partial\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{2z}\left(\mathrm{x}−\mathrm{1}\right)}{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{x}\right)^{\mathrm{3}} } \\ $$ Answered by…

Calculate-the-lattice-enthalpy-of-MgBr-2-Given-that-Enthalpy-of-formation-of-MgBr-2-524-kJ-mol-1-Sublimation-energy-of-Mg-2187-kJ-mol-1-Vaporisation-energy-of-Br-2-l-31-kJ-mol-

Question Number 23694 by Tinkutara last updated on 04/Nov/17 $${Calculate}\:{the}\:{lattice}\:{enthalpy}\:{of}\:{MgBr}_{\mathrm{2}} . \\ $$$${Given}\:{that} \\ $$$${Enthalpy}\:{of}\:{formation}\:{of}\:{MgBr}_{\mathrm{2}} \:=\:−\mathrm{524} \\ $$$${kJ}\:{mol}^{−\mathrm{1}} \\ $$$${Sublimation}\:{energy}\:{of}\:{Mg}\:=\:+\mathrm{2187} \\ $$$${kJ}\:{mol}^{−\mathrm{1}} \\ $$$${Vaporisation}\:{energy}\:{of}\:{Br}_{\mathrm{2}} \left({l}\right)\:=\:+\mathrm{31}…

A-uniform-rope-of-length-L-and-mass-per-unit-having-one-end-fixed-with-the-ceiling-is-released-from-rest-Find-the-tension-in-the-fixed-end-as-a-function-of-the-distance-travelled-by-the-movable-end

Question Number 23666 by Tinkutara last updated on 03/Nov/17 $$\mathrm{A}\:\mathrm{uniform}\:\mathrm{rope}\:\mathrm{of}\:\mathrm{length}\:{L}\:\mathrm{and}\:\mathrm{mass} \\ $$$$\mathrm{per}\:\mathrm{unit}\:\lambda\:\mathrm{having}\:\mathrm{one}\:\mathrm{end}\:\mathrm{fixed}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{ceiling}\:\mathrm{is}\:\mathrm{released}\:\mathrm{from}\:\mathrm{rest}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{fixed}\:\mathrm{end}\:\mathrm{as}\:\mathrm{a}\:\mathrm{function} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{travelled}\:\mathrm{by}\:\mathrm{the}\:\mathrm{movable} \\ $$$$\mathrm{end}. \\ $$ Commented by Tinkutara…

Ethylene-on-combustion-gives-carbon-dioxide-and-water-Its-enthalpy-of-combustion-is-1410-kJ-mol-If-the-enthalpy-of-formation-of-CO-2-and-H-2-O-are-393-3-kJ-and-286-2-kJ-respectively-Calculate-the-

Question Number 23645 by Tinkutara last updated on 03/Nov/17 $$\mathrm{Ethylene}\:\mathrm{on}\:\mathrm{combustion}\:\mathrm{gives}\:\mathrm{carbon} \\ $$$$\mathrm{dioxide}\:\mathrm{and}\:\mathrm{water}.\:\mathrm{Its}\:\mathrm{enthalpy}\:\mathrm{of} \\ $$$$\mathrm{combustion}\:\mathrm{is}\:\mathrm{1410}\:\mathrm{kJ}/\mathrm{mol}.\:\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{enthalpy}\:\mathrm{of}\:\mathrm{formation}\:\mathrm{of}\:\mathrm{CO}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$$$\mathrm{are}\:\mathrm{393}.\mathrm{3}\:\mathrm{kJ}\:\mathrm{and}\:\mathrm{286}.\mathrm{2}\:\mathrm{kJ}\:\mathrm{respectively}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{enthalpy}\:\mathrm{of}\:\mathrm{formation}\:\mathrm{of} \\ $$$$\mathrm{ethylene}. \\…

Question-23632

Question Number 23632 by ajfour last updated on 02/Nov/17 Commented by mrW1 last updated on 03/Nov/17 $$\alpha_{\mathrm{A}} =\frac{\mathrm{d}\omega_{\mathrm{A}} }{\mathrm{dt}}=\frac{\mathrm{d}\omega_{\mathrm{A}} }{\mathrm{d}\theta}×\frac{\mathrm{d}\theta}{\mathrm{dt}}=\frac{\mathrm{d}\omega_{\mathrm{A}} }{\mathrm{d}\theta}×\omega\neq\frac{\mathrm{d}\omega_{\mathrm{A}} }{\mathrm{d}\theta}×\omega_{\mathrm{A}} \\ $$ Commented…

The-system-is-released-from-rest-All-surfaces-are-smooth-Find-the-angle-at-which-the-acceleration-of-wedge-is-maximum-given-M-m-1-2-

Question Number 23617 by Tinkutara last updated on 02/Nov/17 $$\mathrm{The}\:\mathrm{system}\:\mathrm{is}\:\mathrm{released}\:\mathrm{from}\:\mathrm{rest}.\:\mathrm{All} \\ $$$$\mathrm{surfaces}\:\mathrm{are}\:\mathrm{smooth}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\theta\:\mathrm{at} \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{wedge}\:\mathrm{is} \\ $$$$\mathrm{maximum}.\:\left(\mathrm{given}\:\frac{{M}}{{m}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$ Commented by Tinkutara last updated on 02/Nov/17…