Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c

when t = 0, Q = 200 L × 1 g/L = 200 g

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min
Answer:
(1, 9/2) or (1, 4.5)
Step-by-step explanation:
To find the midpoint, take the x1 and x2 values and divide it by 2, same for the y values.
(4-2)/2 and (6 + 3)/2
1 and 4.5
(1, 4.5)
Answer:
The answer is incorrect because 12 should be negative not positive.
Step-by-step explanation:
1. 19 + -31 = - 12
2. Twelve is negative because if you were to start off at -31 and add 19 the answer would remain negative since -31 in a sense is "bigger" than 19. You would need more of a positive number to make -31 positive such as 40.
40+ -31 = 9
Answer:
im nearing the end, of my fourth year, i feel like ive been lacking, crying too many tears, everyone seemed to say, “it was so great!” but did i miss out, was it a huge mistake?
Step-by-step explanation:
i cant help the fact that i like to be alone, it might sound kinda sad, but thats just what i seem to know, i tend to handle things usually by myself, and i cant ever seem to try and ask for help.