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anonymous
 4 years ago
pure water is poured at the rate of 3 gal/min to a tank containing 300 lbs. of salt dissolved in 100 gals. of water and solution kept while stirring, forced out at 2 gals/min. Find the amount of salt at the end of 1 hr. and 30 minutes?
anonymous
 4 years ago
pure water is poured at the rate of 3 gal/min to a tank containing 300 lbs. of salt dissolved in 100 gals. of water and solution kept while stirring, forced out at 2 gals/min. Find the amount of salt at the end of 1 hr. and 30 minutes?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Tank loses 2 gal/min while adding 3 gal/min for a net of 1 gal/min \[Volume = 100 + t\] Salt content in the tank starts at 300 and loses 2C lbs every min where C(t) is the concentration of salt at time t C(t) = Salt content/Volume \[C(t) = \frac{3002t*C(t)}{100+t}\] solving for C(t) \[\rightarrow C(t) = \frac{300}{100+3t}\] Let S(t) be salt content at time t S(t) = C(t)*Volume \[\rightarrow \ s(t) = \frac{300(100+t)}{100+3t}\] when t=90 \[s(90) = \frac{300(190)}{100+270} = 203.57\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks for the effort dumb cow, but i already got the answer. . it's 83.10 lbs.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can you please explain to me how you got that answer? Thanks

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this problem is actually related to mixture problem related to differential equation application. . have you encountered this one?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0here, let me show you how i got it. .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328684363825:dw we got 100 +t as denominator for the because the rate of going in minus the rate in going out (3t  2t) resulted to t. .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0to make the equation\[dQ/dt=3(0)  2(Q/100+t)\] we're finding the change of concentration

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328684946025:dwdw:1328685032254:dw cancel ln.dw:1328685126662:dw this is now our equation. .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now let's find the value for C or constant, given the initial number of salt (Q) which is 300 at t=0 dw:1328685267814:dw now that we have the value for constant, we can now find the remaining salt in lbs at t=90 mins.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328685463206:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's how i got the answer. . kindly check if there are irregularities.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No it looks fine, thanks for the explanation. i didn't see where i could solve dQ/dt explicitly like that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0aah, yea me either got difficulties in solving the problem. but anyways, thanks for the medal.
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