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Klio2033 [76]
3 years ago
5

What places are likely to support a large producer population?

Physics
2 answers:
Lady_Fox [76]3 years ago
8 0
The answer is B.
because most of  organism need water and sunlight to support their live and their growth
max2010maxim [7]3 years ago
7 0
In general, producer populations tend to thrive in areas with lots of water and light, so the answer would be <span>"B: Areas with abundant water and sunlight."</span> This of course isn't always the case, however.
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A tank contains 150 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
nikitadnepr [17]

Answer:

The number A(t) of grams of salt in the tank at time t is A(t) = 150 - 110 e^{-\frac{t}{50} }

Explanation:

Knowing

\frac{dA}{dt} = Rin - Rout

First we have to find the Rin and Rout

Rin = (concentration of the salt inflow) * (input rate of brine)

Rin = 1 g/L * 3 L/min = 3 g/L

Rout = (concentration of the salt outflow) * (output rate of brine)

Rout = (\frac{A(t)}{150} g/L) * (3 L/min) = \frac{A(t)}{50} g/min

Substituting this results

\frac{dA}{dt} = 3 - \frac{A(t)}{50} --> \frac{dA}{dt} + \frac{1}{50} A(t)  = 4

Thus, integration factors is

e^{ \int\limits^._. {\frac{1}{50} } \, dt } = e^{\frac{t}{50} }

e^{\frac{t}{50} } \frac{dA}{dt} + \frac{1}{50} e^{\frac{t}{50} A(t) = 4 e^{\frac{t}{50} }

e^{\frac{t}{50} } A(t) = \int\limits^._. {3 e^{\frac{t}{50} } } \, dt\\ \\A(t) = 150 + c e^{-\frac{t}{50} }

Applying the initial conditions

A(0) = 40

c = 150 - 40 = 110

Now, substitute this result in the solution to get

A(t) = 150 - 110 e^{-\frac{t}{50} }

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