A tank contains 300 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.
Answers
Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
___
The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
The number of grams of salt in the tank at any time t is 40 grams. The inflow and outflow of brine do not affect the amount of salt in the tank because the solution is well-mixed, and the salt concentration remains constant.
To solve this problem, we need to set up a differential equation that describes the rate of change of salt in the tank over time. Let A(t) represent the number of grams of salt in the tank at time t.
Let's break down the components affecting the rate of change of salt in the tank:
Salt inflow rate: The brine is being pumped into the tank at a constant rate of 6 liters per minute, and it contains 1 gram of salt per liter. So, the rate of salt inflow is 6 grams per minute.
Salt outflow rate: The solution in the tank is being pumped out at the same rate of 6 liters per minute, which means the rate of salt outflow is also 6 grams per minute.
Mixing of the solution: Since the tank is well-mixed, the concentration of salt remains uniform throughout the tank.
Now, let's set up the differential equation for A(t):
dA/dt = Rate of salt inflow - Rate of salt outflow
dA/dt = 6 grams/min - 6 grams/min
dA/dt = 0
The above equation shows that the rate of change of salt in the tank is constant and equal to zero. This means the number of grams of salt in the tank remains constant over time.
Now, let's find the constant value of A(t) using the initial condition where the tank initially contains 40 grams of salt.
When t = 0, A(0) = 40 grams
Since the rate of change is zero, A(t) will be the same as the initial amount of salt in the tank at any time t:
A(t) = 40 grams
So, the number of grams of salt in the tank at any time t is 40 grams. The inflow and outflow of brine do not affect the amount of salt in the tank because the solution is well-mixed, and the salt concentration remains constant.
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Answer:
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Step-by-step explanation:
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rate of growth of a population with a birth rate of 30births per
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Answers
Answer:
10 growth per 1000.
Step-by-step explanation:
Given,
Rate of birth = 30 births per 1000
Rate of death = 20 deaths per 1000
As the growth in population is the difference in the number of the child take birth and the person die.
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Answers
Answer:
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Step-by-step explanation:
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5 and 2 are opposite angles and they are connected by, technically, two lines crossing each other.
PLS HELP ME!!!
Answers
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Answers
Answer: 1.88
Step-by-step explanation:
The confidence interval for population mean is given by :-
Given : Significance level :
Critical value :
By using the standard normal distribution table for z, we find the critical value of corresponds to the p-value 0.03.
Hence, the z-value that was used in the computation must be 1.88 .
Answers
hope this is helpful