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Question Number 74266 by smartsmith459@gmail.com last updated on 22/Nov/19
please kindly help me with the solutions to these question?very urgent please (1) if dy=x^3 dx. find the equation of y in terms of x   if the curve passes through (1,1)  2) Given that the volume v(t) of cell at a time t changes according to  ((dV(t))/dt)= sin t, with v(t)=4. find v(t)  3) Given (dP/dt) + 3P = 0. determine P (t)   if p(0)= 4  4) Radium decomposes at a rate proportion to the amount   present. if the half−life of the radium is  1000 years. what is the percentage lost in 100 years?  5) Calculate the pressure of a gas after phase  transition at 171°K from 101.3kPa  pressure at 472°K taking R = 0.1886kJ\kgK and L = 35.73kg
$${please}\:{kindly}\:{help}\:{me}\:{with}\:{the}\:{solutions}\:{to}\:{these}\:{question}?{very}\:{urgent}\:{please}\:\left(\mathrm{1}\right)\:{if}\:{dy}={x}^{\mathrm{3}} {dx}.\:{find}\:{the}\:{equation}\:{of}\:{y}\:{in}\:{terms}\:{of}\:{x} \\ $$$$\:{if}\:{the}\:{curve}\:{passes}\:{through}\:\left(\mathrm{1},\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:{Given}\:{that}\:{the}\:{volume}\:{v}\left({t}\right)\:{of}\:{cell}\:{at}\:{a}\:{time}\:{t}\:{changes}\:{according}\:{to} \\ $$$$\frac{{dV}\left({t}\right)}{{dt}}=\:{sin}\:{t},\:{with}\:{v}\left({t}\right)=\mathrm{4}.\:{find}\:{v}\left({t}\right) \\ $$$$\left.\mathrm{3}\right)\:{Given}\:\frac{{dP}}{{dt}}\:+\:\mathrm{3}{P}\:=\:\mathrm{0}.\:{determine}\:{P}\:\left({t}\right)\: \\ $$$${if}\:{p}\left(\mathrm{0}\right)=\:\mathrm{4} \\ $$$$\left.\mathrm{4}\right)\:{Radium}\:{decomposes}\:{at}\:{a}\:{rate}\:{proportion}\:{to}\:{the}\:{amount}\: \\ $$$${present}.\:{if}\:{the}\:{half}−{life}\:{of}\:{the}\:{radium}\:{is} \\ $$$$\mathrm{1000}\:{years}.\:{what}\:{is}\:{the}\:{percentage}\:{lost}\:{in}\:\mathrm{100}\:{years}? \\ $$$$\left.\mathrm{5}\right)\:{Calculate}\:{the}\:{pressure}\:{of}\:{a}\:{gas}\:{after}\:{phase} \\ $$$${transition}\:{at}\:\mathrm{171}°{K}\:{from}\:\mathrm{101}.\mathrm{3}{kPa} \\ $$$${pressure}\:{at}\:\mathrm{472}°{K}\:{taking}\:{R}\:=\:\mathrm{0}.\mathrm{1886}{kJ}\backslash{kgK}\:{and}\:{L}\:=\:\mathrm{35}.\mathrm{73}{kg} \\ $$

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