A tank contains 100L of water. A solution with a salt concentration of 0.6kg/L i

Question

A tank contains 100L of water. A solution with a salt concentration of 0.6kg/L is added at a rate of 7L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 5L/min. Answer the following questions. 1. If is the amount of salt (in kilograms) after t minutes, what is the differential equation for which y is satisfied? Use the variable y for y(t). Answer (in kilograms per minute): dy/dt = 4.2-(5y/100+2t) 2. How mu Ch salt is in the tank after 40 minutes? Answer (in kilograms):

Explanation / Answer

Adding salt solution = 0.6 x 7 x t

removing = 5 x t x concentration

concentration = 0.6 x 7 x t/(100+(7 x t))

dy/dt = (0.6 x 7 x t) - [5 x t x {0.6 x 7 x t/(100+(7 x t))}]

After 40 min the answer is

(0.6 x 7 x 40) - [5 x 40 x {0.6 x 7 x 40/(100+(7 x 40)}]

= 79.6 Kg

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