Question Number 124976 by liberty last updated on 07/Dec/20
$$\left(\mathrm{1}\right)\:{The}\:{gravitational}\:{force}\:\left({in}\:{lb}\right)\:{of} \\ $$$${attraction}\:{between}\:{two}\:{objects}\:{is}\:{given} \\ $$$${by}\:{F}\:=\frac{{k}}{{x}^{\mathrm{2}} },\:{where}\:{x}\:{is}\:{the}\:{distance} \\ $$$${between}\:{the}\:{objects}.\:{If}\:{the}\:{objects}\:{are} \\ $$$$\mathrm{10}\:{ft}\:{apart},\:{find}\:{the}\:{work}\:{required}\:{to} \\ $$$${separate}\:{them}\:{until}\:{they}\:{are}\:\mathrm{50}\:{ft}\:{apart}.\:{Express} \\ $$$${the}\:{result}\:{in}\:{terms}\:{of}\:{k}. \\ $$$$\left({a}\right)\:\frac{{k}}{\mathrm{500}}\:\:\:\:\:\:\left({b}\right)\:\frac{\mathrm{2}{k}}{\mathrm{25}}\:\:\:\:\:\left({c}\right)\:\frac{{k}}{\mathrm{5}}\:\:\:\left({d}\right)\:\frac{{k}}{\mathrm{40}} \\ $$$$\left(\mathrm{2}\right){One}\:{end}\:{of}\:{a}\:{pool}\:{is}\:{vertical}\:{wall}\:\mathrm{15}\:{ft} \\ $$$${wide}.\:{What}\:{is}\:{the}\:{force}\:{exerted}\:{on}\:{this} \\ $$$${wall}\:{by}\:{the}\:{water}\:{if}\:{it}\:{is}\:\mathrm{6}\:{ft}\:{deep}? \\ $$$${The}\:{density}\:{of}\:{water}\:{is}\:\mathrm{62}.\mathrm{4}\:{lb}/{ft}^{\mathrm{3}} \\ $$$$\left({a}\right)\:\mathrm{8420}\:{lb}\:\:\:\:\left({b}\right)\:\mathrm{33},\mathrm{700}\:{lb}\:\:\:\:\left({c}\right)\:\mathrm{2810}\:{lb}\:\:\left({d}\right)\:\mathrm{16},\mathrm{800}\:{lb} \\ $$$$\left(\mathrm{3}\right){Find}\:{the}\:{area}\:{of}\:{the}\:{surface}\:{generated} \\ $$$${by}\:{revolving}\:{the}\:{curve}\:{about}\:{that}\: \\ $$$${indicated}\:{axis}.\:\:{x}\:=\:\mathrm{3}\sqrt{\mathrm{4}−{y}}\:,\:\mathrm{0}\leqslant{y}\leqslant\frac{\mathrm{15}}{\mathrm{4}}\:,\:{y}−{axis} \\ $$$$\left({a}\right)\:\left(\frac{\mathrm{125}}{\mathrm{2}}+\mathrm{5}\sqrt{\mathrm{10}}\right)\pi\:\:\:\:\:\:\:\left({b}\right)\:\left(\frac{\mathrm{125}}{\mathrm{2}}−\mathrm{5}\sqrt{\mathrm{10}}\right)\pi \\ $$$$\left({c}\right)\:\frac{\mathrm{125}}{\mathrm{2}}\pi\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({d}\right)\:\mathrm{5}\pi\sqrt{\mathrm{10}}\: \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 07/Dec/20
![Work done =∫_(10) ^(50) (k/x^2 )dx=[−(k/x)]_(10) ^(50) =((2k)/(25))](https://www.tinkutara.com/question/Q124977.png)
$${Work}\:{done}\:=\int_{\mathrm{10}} ^{\mathrm{50}} \frac{{k}}{{x}^{\mathrm{2}} }{dx}=\left[−\frac{{k}}{{x}}\right]_{\mathrm{10}} ^{\mathrm{50}} =\frac{\mathrm{2}{k}}{\mathrm{25}} \\ $$
Answered by bemath last updated on 07/Dec/20
![(3)S = ∫_0 ^(15/4) 2πx (√(1+((dx/dy))^2 )) dy x = 3(√(4−y)) ; (dx/dy) = ((−3)/(2(√(4−y)))) then 1+((dx/dy))^2 =1+(9/(4(4−y)))=((25−4y)/(4(4−y))) S = ∫_0 ^( ((15)/4)) 6π(√(4−y)) (√((25−4y)/(4(4−y)))) dy S= 3π∫_0 ^((15)/4) (√(25−4y)) dy = [(1/((3/2)(−4)))×3π×(√((25−4y)^3 )) ]_0 ^((15)/4) = −(π/2) {(√(1000))−125 } = ((125π)/2)−5π(√(10)) = (((125)/2)−5(√(10)))π](https://www.tinkutara.com/question/Q124981.png)
$$\left(\mathrm{3}\right){S}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{15}/\mathrm{4}} {\int}}\mathrm{2}\pi{x}\:\sqrt{\mathrm{1}+\left(\frac{{dx}}{{dy}}\right)^{\mathrm{2}} }\:{dy} \\ $$$${x}\:=\:\mathrm{3}\sqrt{\mathrm{4}−{y}}\:;\:\frac{{dx}}{{dy}}\:=\:\frac{−\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{4}−{y}}} \\ $$$$\:{then}\:\mathrm{1}+\left(\frac{{dx}}{{dy}}\right)^{\mathrm{2}} =\mathrm{1}+\frac{\mathrm{9}}{\mathrm{4}\left(\mathrm{4}−{y}\right)}=\frac{\mathrm{25}−\mathrm{4}{y}}{\mathrm{4}\left(\mathrm{4}−{y}\right)} \\ $$$${S}\:=\:\underset{\mathrm{0}} {\overset{\:\frac{\mathrm{15}}{\mathrm{4}}} {\int}}\:\mathrm{6}\pi\sqrt{\mathrm{4}−{y}}\:\sqrt{\frac{\mathrm{25}−\mathrm{4}{y}}{\mathrm{4}\left(\mathrm{4}−{y}\right)}}\:{dy} \\ $$$${S}=\:\mathrm{3}\pi\underset{\mathrm{0}} {\overset{\frac{\mathrm{15}}{\mathrm{4}}} {\int}}\sqrt{\mathrm{25}−\mathrm{4}{y}}\:\:{dy}\:=\:\left[\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{2}}\left(−\mathrm{4}\right)}×\mathrm{3}\pi×\sqrt{\left(\mathrm{25}−\mathrm{4}{y}\right)^{\mathrm{3}} }\:\right]_{\mathrm{0}} ^{\frac{\mathrm{15}}{\mathrm{4}}} \\ $$$$=\:−\frac{\pi}{\mathrm{2}}\:\left\{\sqrt{\mathrm{1000}}−\mathrm{125}\:\right\} \\ $$$$=\:\frac{\mathrm{125}\pi}{\mathrm{2}}−\mathrm{5}\pi\sqrt{\mathrm{10}}\:=\:\left(\frac{\mathrm{125}}{\mathrm{2}}−\mathrm{5}\sqrt{\mathrm{10}}\right)\pi \\ $$
Answered by bemath last updated on 07/Dec/20
$$\left(\mathrm{2}\right){Force}\:=\:{whA}\:=\:\mathrm{62}.\mathrm{4}\:{lb}/{ft}^{\mathrm{3}} \:×\mathrm{6}\:{ft}\:×\mathrm{15}^{\mathrm{2}} \:{ft}^{\mathrm{2}} \\ $$$$\:=\:\mathrm{84},\mathrm{240}\:{lb} \\ $$