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wariber [46]
4 years ago
13

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 L of a dye solution with a

concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 L/min, the well-stirred solution flowing out at the same rate.
a. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Mathematics
1 answer:
In-s [12.5K]4 years ago
4 0

Answer: t= 460.52 minutes

Step-by-step explanation:Q'=Q/100

Q'= rate in and out of water

Finding the differential equation

Let Q'(t)= The quantity of dye in the tank for t time

But rate in=0 Q/200 ×2=Q'

Q'/Q=-1/100

Dividing by Q gives

Ln/Q/ + c = -1/100 + c1

Integrating both sides gives

Ln/Q/ = -(1/100)t + c2

But c+c1=C2= A constant

Q=C2e(-t/100)

200e-(t/100)

t= ln200

t=460.52minutes

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0.3% = 0.003..." of " means multiply
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0.85% = 0.0085
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what u can do to turn the percents to decimals is divide by 100.
0.3 / 100 = 0.003 and 0.85 / 100 = 0.0085
or u can move the decimal two spaces to the left
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