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A rescue cable attached to a helicopter′s weighs 2 lb/ft. A man 180−lb grabs the end of the rope and his pulled from the ocean into the helicopter. How much work is done in lifting the man if the helicopter is 40 ft above the water ? (a) 8800 lb−ft (b) 1780 lb−ft (c) 7280 lb−ft (d) 10,400 lb−ft

$$\:{A}\:{rescue}\:{cable}\:{attached}\:{to}\:{a}\: \\ $$$${helicopter}'{s}\:{weighs}\:\mathrm{2}\:{lb}/{ft}.\: \\ $$$${A}\:{man}\:\mathrm{180}−{lb}\:{grabs}\:{the}\:{end}\: \\ $$$${of}\:{the}\:{rope}\:{and}\:{his}\:{pulled}\: \\ $$$${from}\:{the}\:{ocean}\:{into}\:{the}\:{helicopter}. \\ $$$${How}\:{much}\:{work}\:{is}\:{done}\:{in}\: \\ $$$${lifting}\:{the}\:{man}\:{if}\:{the}\:{helicopter} \\ $$$${is}\:\mathrm{40}\:{ft}\:{above}\:{the}\:{water}\:? \\ $$$$\left({a}\right)\:\mathrm{8800}\:{lb}−{ft} \\ $$$$\left({b}\right)\:\mathrm{1780}\:{lb}−{ft} \\ $$$$\left({c}\right)\:\mathrm{7280}\:{lb}−{ft} \\ $$$$\left({d}\right)\:\mathrm{10},\mathrm{400}\:{lb}−{ft} \\ $$

Question Number 125050 Answers: 1 Comments: 0

∫_0 ^π (e^(cos x) /(e^(cos x) +e^(−cos x) )) dx =?

$$\:\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{{e}^{\mathrm{cos}\:{x}} }{{e}^{\mathrm{cos}\:{x}} +{e}^{−\mathrm{cos}\:{x}} }\:{dx}\:=?\: \\ $$

Question Number 125048 Answers: 0 Comments: 1

It takes a force of 19,000 lb to compress a spring from its free height of 15 in to its fully compressed height of 10 in. How much work does it take to compress the spring the first in? (a) 1900 in.−lb (b) 950 in.−lb (c) 3800 in.−lb (d) 190,000 in.−lb

$${It}\:{takes}\:{a}\:{force}\:{of}\:\mathrm{19},\mathrm{000}\:{lb}\:{to} \\ $$$${compress}\:{a}\:{spring}\:{from}\:{its}\:{free} \\ $$$${height}\:{of}\:\mathrm{15}\:{in}\:{to}\:{its}\:{fully}\: \\ $$$${compressed}\:{height}\:{of}\:\mathrm{10}\:{in}.\:{How} \\ $$$${much}\:\:{work}\:{does}\:{it}\:{take}\:{to}\: \\ $$$${compress}\:{the}\:{spring}\:{the}\:{first}\:{in}? \\ $$$$\left({a}\right)\:\mathrm{1900}\:{in}.−{lb} \\ $$$$\left({b}\right)\:\mathrm{950}\:{in}.−{lb} \\ $$$$\left({c}\right)\:\mathrm{3800}\:{in}.−{lb} \\ $$$$\left({d}\right)\:\mathrm{190},\mathrm{000}\:{in}.−{lb} \\ $$

Question Number 125034 Answers: 0 Comments: 1

find ∫_0 ^(π/2) (x/(sinx))dx

$$\mathrm{find}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{x}}{\mathrm{sinx}}\mathrm{dx} \\ $$

Question Number 124985 Answers: 1 Comments: 3

Question Number 124980 Answers: 1 Comments: 0

find ∫ (dx/(((√(x−1))+2(√(x+1)))^2 ))

$$\mathrm{find}\:\int\:\:\frac{\mathrm{dx}}{\left(\sqrt{\mathrm{x}−\mathrm{1}}+\mathrm{2}\sqrt{\mathrm{x}+\mathrm{1}}\right)^{\mathrm{2}} } \\ $$

Question Number 124979 Answers: 4 Comments: 0

let f(x)=arctan(x^n ) with n natural 1) find f^((n)) (0) and f^((n)) (1) 2)developp f at integr serie 3)calculte ∫_0 ^∞ ((f(x))/x^n )dx with n≥2

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{arctan}\left(\mathrm{x}^{\mathrm{n}} \right)\:\mathrm{with}\:\mathrm{n}\:\mathrm{natural} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{find}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right)\:\mathrm{and}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie} \\ $$$$\left.\mathrm{3}\right)\mathrm{calculte}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{n}} }\mathrm{dx}\:\mathrm{with}\:\mathrm{n}\geqslant\mathrm{2} \\ $$

Question Number 124976 Answers: 2 Comments: 1

(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))

$$\left(\mathrm{1}\right)\:{The}\:{gravitational}\:{force}\:\left({in}\:{lb}\right)\:{of} \\ $$$${attraction}\:{between}\:{two}\:{objects}\:{is}\:{given} \\ $$$${by}\:{F}\:=\frac{{k}}{{x}^{\mathrm{2}} },\:{where}\:{x}\:{is}\:{the}\:{distance} \\ $$$${between}\:{the}\:{objects}.\:{If}\:{the}\:{objects}\:{are} \\ $$$$\mathrm{10}\:{ft}\:{apart},\:{find}\:{the}\:{work}\:{required}\:{to} \\ $$$${separate}\:{them}\:{until}\:{they}\:{are}\:\mathrm{50}\:{ft}\:{apart}.\:{Express} \\ $$$${the}\:{result}\:{in}\:{terms}\:{of}\:{k}. \\ $$$$\left({a}\right)\:\frac{{k}}{\mathrm{500}}\:\:\:\:\:\:\left({b}\right)\:\frac{\mathrm{2}{k}}{\mathrm{25}}\:\:\:\:\:\left({c}\right)\:\frac{{k}}{\mathrm{5}}\:\:\:\left({d}\right)\:\frac{{k}}{\mathrm{40}} \\ $$$$\left(\mathrm{2}\right){One}\:{end}\:{of}\:{a}\:{pool}\:{is}\:{vertical}\:{wall}\:\mathrm{15}\:{ft} \\ $$$${wide}.\:{What}\:{is}\:{the}\:{force}\:{exerted}\:{on}\:{this} \\ $$$${wall}\:{by}\:{the}\:{water}\:{if}\:{it}\:{is}\:\mathrm{6}\:{ft}\:{deep}? \\ $$$${The}\:{density}\:{of}\:{water}\:{is}\:\mathrm{62}.\mathrm{4}\:{lb}/{ft}^{\mathrm{3}} \\ $$$$\left({a}\right)\:\mathrm{8420}\:{lb}\:\:\:\:\left({b}\right)\:\mathrm{33},\mathrm{700}\:{lb}\:\:\:\:\left({c}\right)\:\mathrm{2810}\:{lb}\:\:\left({d}\right)\:\mathrm{16},\mathrm{800}\:{lb} \\ $$$$\left(\mathrm{3}\right){Find}\:{the}\:{area}\:{of}\:{the}\:{surface}\:{generated} \\ $$$${by}\:{revolving}\:{the}\:{curve}\:{about}\:{that}\: \\ $$$${indicated}\:{axis}.\:\:{x}\:=\:\mathrm{3}\sqrt{\mathrm{4}−{y}}\:,\:\mathrm{0}\leqslant{y}\leqslant\frac{\mathrm{15}}{\mathrm{4}}\:,\:{y}−{axis} \\ $$$$\left({a}\right)\:\left(\frac{\mathrm{125}}{\mathrm{2}}+\mathrm{5}\sqrt{\mathrm{10}}\right)\pi\:\:\:\:\:\:\:\left({b}\right)\:\left(\frac{\mathrm{125}}{\mathrm{2}}−\mathrm{5}\sqrt{\mathrm{10}}\right)\pi \\ $$$$\left({c}\right)\:\frac{\mathrm{125}}{\mathrm{2}}\pi\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({d}\right)\:\mathrm{5}\pi\sqrt{\mathrm{10}}\: \\ $$$$ \\ $$

Question Number 124957 Answers: 1 Comments: 1