(1) The gravitational force (in lb) of attraction between two objects is given by F =(k/x^2 ), where x is the distance between the objects. If the objects are 10 ft apart, find the work required to separate them until they are 50 ft apart. Express the result in terms of k. (a) (k/(500)) (b) ((2k)/(25)) (c) (k/5) (d) (k/(40)) (2)One end of a pool is vertical wall 15 ft wide. What is the force exerted on this wall by the water if it is 6 ft deep? The density of water is 62.4 lb/ft^3 (a) 8420 lb (b) 33,700 lb (c) 2810 lb (d) 16,800 lb (3)Find the area of the surface generated by revolving the curve about that indicated axis. x = 3(√(4−y)) , 0≤y≤((15)/4) , y−axis (a) (((125)/2)+5(√(10)))π (b) (((125)/2)−5(√(10)))π (c) ((125)/2)π (d) 5π(√(10))


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Question Number 124976 by liberty last updated on 07/Dec/20

$(1) T h e g r a v i t a t i o n a l f o r c e (i n l b) o f$ $a t t r a c t i o n b e t w e e n t w o o b j e c t s i s g i v e n$ $b y F = \frac{k}{x^{2}}, w h e r e x i s t h e d i s t a n c e$ $b e t w e e n t h e o b j e c t s . I f t h e o b j e c t s a r e$ $10 f t a p a r t, f i n d t h e w o r k r e q u i r e d t o$ $s e p a r a t e t h e m u n t i l t h e y a r e 50 f t a p a r t . E x p r e s s$ $t h e r e s u l t i n t e r m s o f k .$ $(a) \frac{k}{500} (b) \frac{2 k}{25} (c) \frac{k}{5} (d) \frac{k}{40}$ $(2) O n e e n d o f a p o o l i s v e r t i c a l w a l l 15 f t$ $w i d e . W h a t i s t h e f o r c e e x e r t e d o n t h i s$ $w a l l b y t h e w a t e r i f i t i s 6 f t d e e p ?$ $T h e d e n s i t y o f w a t e r i s 62.4 l b / {f t}^{3}$ $(a) 8420 l b (b) 33, 700 l b (c) 2810 l b (d) 16, 800 l b$ $(3) F i n d t h e a r e a o f t h e s u r f a c e g e n e r a t e d$ $b y r e v o l v i n g t h e c u r v e a b o u t t h a t$ $i n d i c a t e d a x i s . x = 3 \sqrt{4 - y}, 0 ⩽ y ⩽ \frac{15}{4}, y - a x i s$ $(a) (\frac{125}{2} + 5 \sqrt{10}) π (b) (\frac{125}{2} - 5 \sqrt{10}) π$ $(c) \frac{125}{2} π (d) 5 π \sqrt{10}$
Commented by Dwaipayan Shikari last updated on 07/Dec/20

$W o r k d o n e = \int_{10}^{50} \frac{k}{x^{2}} d x = {[- \frac{k}{x}]}_{10}^{50} = \frac{2 k}{25}$
	Answered by bemath last updated on 07/Dec/20
	$(3)S = ∫_0 ^(15/4) 2πx (√(1+((dx/dy))^2 )) dy x = 3(√(4−y)) ; (dx/dy) = ((−3)/(2(√(4−y)))) then 1+((dx/dy))^2 =1+(9/(4(4−y)))=((25−4y)/(4(4−y))) S = ∫_0 ^( ((15)/4)) 6π(√(4−y)) (√((25−4y)/(4(4−y)))) dy S= 3π∫_0 ^((15)/4) (√(25−4y)) dy = [(1/((3/2)(−4)))×3π×(√((25−4y)^3 )) ]_0 ^((15)/4) = −(π/2) {(√(1000))−125 } = ((125π)/2)−5π(√(10)) = (((125)/2)−5(√(10)))π$
	$(3) S = \underset{0}{\int^{15 / 4}} 2 π x \sqrt{1 + {(\frac{d x}{d y})}^{2}} d y$ $x = 3 \sqrt{4 - y}; \frac{d x}{d y} = \frac{- 3}{2 \sqrt{4 - y}}$ $t h e n 1 + {(\frac{d x}{d y})}^{2} = 1 + \frac{9}{4 (4 - y)} = \frac{25 - 4 y}{4 (4 - y)}$ $S = \underset{0}{\int^{\frac{15}{4}}} 6 π \sqrt{4 - y} \sqrt{\frac{25 - 4 y}{4 (4 - y)}} d y$ $S = 3 π \underset{0}{\int^{\frac{15}{4}}} \sqrt{25 - 4 y} d y = {[\frac{1}{\frac{3}{2} (- 4)} \times 3 π \times \sqrt{{(25 - 4 y)}^{3}}]}_{0}^{\frac{15}{4}}$ $= - \frac{π}{2} {\sqrt{1000} - 125}$ $= \frac{125 π}{2} - 5 π \sqrt{10} = (\frac{125}{2} - 5 \sqrt{10}) π$
	Answered by bemath last updated on 07/Dec/20

	$(2) F o r c e = w h A = 62.4 l b / {f t}^{3} \times 6 f t \times 15^{2} {f t}^{2}$ $= 84, 240 l b$
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