Question Number 125999 by bramlexs22 last updated on 16/Dec/20 | ||
Commented by bramlexs22 last updated on 16/Dec/20 | ||
$${thanks}\:{both}\:{sir} \\ $$ | ||
Answered by mr W last updated on 16/Dec/20 | ||
$${V}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×\mathrm{4}×\mathrm{10}=\mathrm{80}\:{m}^{\mathrm{3}} \\ $$$${W}=\mathrm{80}×\mathrm{1000}×\mathrm{9}.\mathrm{8}×\left(\mathrm{3}+\frac{\mathrm{4}}{\mathrm{3}}\right)=\mathrm{3}\:\mathrm{397}\:\mathrm{333}\:{J} \\ $$ | ||
Commented by bramlexs22 last updated on 16/Dec/20 | ||
$${by}\:{integral}\:{sir}? \\ $$ | ||
Commented by mr W last updated on 16/Dec/20 | ||
$${dW}=\rho{gzdV} \\ $$$${W}=\rho{g}\int{zdV}=\rho{gVz}_{{S}} \\ $$$${with}\:{z}_{{S}} ={distance}\:{to}\:{mass}\:{center} \\ $$$${this}\:{is}\:{valid}\:{for}\:{any}\:{shape}\:{of}\:{the} \\ $$$${water}\:{body}\:{to}\:{move}! \\ $$ | ||
Answered by liberty last updated on 16/Dec/20 | ||
$${to}\:{find}\:{the}\:{weight}\:{of}\:{water}\:,\:{multiply}\: \\ $$$${the}\:{acceleration}\:{due}\:{to}\:{gravity}\:,\:{by}\:{the}\: \\ $$$${mass}\:{which}\:{equals}\:{to}\:{density}\:{of}\:{water} \\ $$$${multiplied}\:{by}\:{the}\:{volume}\::\: \\ $$$$\:\:\left(\mathrm{9}.\mathrm{8}\:{m}/{s}^{\mathrm{2}} \right)\left(\mathrm{1},\mathrm{000}\:{kg}/{m}^{\mathrm{3}} \right)\left({x}_{{i}} ^{\ast} \:{m}\right)\left(\Delta{x}\:{m}\right)= \\ $$$$\:\:\mathrm{98},\mathrm{000}\left({x}_{{i}} ^{\ast} \right)\left(\Delta{x}\right)\:{N} \\ $$$$\:\:\Leftrightarrow\:{W}=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{98},\mathrm{000}\right)\left({x}_{{i}} ^{\ast} \right)\left(\mathrm{7}−{x}_{{i}} ^{\ast} \right)\Delta{x} \\ $$$$\Leftrightarrow\:{W}\:=\:\mathrm{98},\mathrm{000}\:\int_{\mathrm{0}} ^{\:\mathrm{4}} {x}\left(\mathrm{7}−{x}\right)\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{98},\mathrm{000}\:\left[\:\frac{\mathrm{7}}{\mathrm{2}}{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \:\right]_{\mathrm{0}} ^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{98},\mathrm{000}\:\left(\frac{\mathrm{7}}{\mathrm{2}}\mathrm{4}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}\mathrm{4}^{\mathrm{3}} \right)=\frac{\mathrm{10},\mathrm{192},\mathrm{000}}{\mathrm{3}}\:{J} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{3},\mathrm{397},\mathrm{333}\:{J}\: \\ $$$$ \\ $$ | ||