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A newly constructed fish pond contains 3000 liters of water. Unfortunately the pond has been contaminated with 5 kg of a toxic chemical during the construction process. The pond's filtering system removes water from the pond at a rate of 200 liters/minute, removes 50% of the chemical, and returns the same volume of (the now somewhat less contaminated) water to the pond. Write a differential equation for the time (measured in minutes) evolution of:

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lublana

Answer:

[tex]\frac{dm}{dt}=-\frac{m}{30}[/tex]

Step-by-step explanation:

We are given that

Volume of water in fish pond,V=3000 L

Mass of toxic  chemical=5 kg

Rate=200L/min

Removes chemical=50%

We have to find the differential equation for the time.

Let mass of toxic at any time t= m kg

Rate=Rate in-Rate out

[tex]\frac{dm}{dt}=200\times 0-\frac{m}{3000}\times 200\times \frac{50}{100}[/tex]

[tex]\frac{dm}{dt}=0-\frac{m}{30}[/tex]

[tex]\frac{dm}{dt}=-\frac{m}{30}[/tex]

MrRoyal

The differential equation for the time is [tex]\frac{dm}{dt} = -\frac{m}{30}[/tex]

The given parameters are:

  • Volume of water in the pond, [tex]V=3000 L[/tex]  
  • Mass of the chemical, [tex]m =5 kg[/tex]
  • Rate, [tex]r = 200L/min[/tex]
  • Proportion of chemicals removed, [tex]p=50\%[/tex]

The differential equation for the time is then calculated as:

[tex]\frac{dm}{dt} = R_{in} -R_{out}[/tex]

Where:

[tex]R_{in} = 200 \times 0 = 0[/tex] --- the rate in

[tex].R_{out} = \frac{m}{3000} \times 200 \times 50\% = \frac{m}{30}[/tex] --- the rate out

Substitute these values in [tex]\frac{dm}{dt} = R_{in} -R_{out}[/tex]

[tex]\frac{dm}{dt} = 0 -\frac{m}{30}[/tex]

[tex]\frac{dm}{dt} = -\frac{m}{30}[/tex]

Hence, the differential equation for the time is [tex]\frac{dm}{dt} = -\frac{m}{30}[/tex]

Read more about differential equations at:

https://brainly.com/question/14763154

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