Answer the question in my attachment

A tank contains 5,000 L of brine with 13 kg of dissolved salt. Pure water enters the tank at a rate of 50 L/min. The solution is

tresset_1 [31]

Answer:

a) $x(t) = 13*e^(^-^\frac{t}{100}^)$

b) 10.643 kg

Step-by-step explanation:

Solution:-

- We will first denote the amount of salt in the solution as x ( t ) at any time t.

- We are given that the Pure water enters the tank ( contains zero salt ).

- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min

- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.

- The ODE is mathematically expressed as:

$\frac{dx}{dt} =$ ( salt flow in ) - ( salt flow out )

- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0

- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).

- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.

- So any time ( t ) the concentration of salt in the 5,000 L is:

$conc = \frac{x(t)}{1000}\frac{kg}{L}$

- The amount of salt leaving the tank per unit time can be determined from:

salt flow-out = conc * V( flow-out )

salt flow-out = $\frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}\\$

salt flow-out = $\frac{x(t)}{100}\frac{kg}{min}$

- The ODE becomes:

$\frac{dx}{dt} = 0 - \frac{x}{100}$

- Separate the variables and integrate both sides:

$\int {\frac{1}{x} } \, dx = -\int\limits^t_0 {\frac{1}{100} } \, dt + c\\\\Ln( x ) = -\frac{t}{100} + c\\\\x = C*e^(^-^\frac{t}{100}^)$

- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:

$13 = C*e^0 = C$

- The solution to the ODE becomes:

$x(t) = 13*e^(^-^\frac{t}{100}^)$

- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:

$x(20) = 13*e^(^-^\frac{20}{100}^)\\\\x(20) = 13*e^(^-^\frac{1}{5}^)\\\\x(20) = 10.643 kg$

- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg

7 0

3 years ago

The sketch of this crayon represents an irregular polygon. Find the area of the figure. 21.75 cm²

hjlf

75 cm that should be it

5 0

3 years ago

GEOMETRY. Name a Pair of Parallel AND PERPENDICULAR lines.

trasher [3.6K]

Answer:

I'm certain that it's AB and HI

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Find the equation of the lines parallel and perpendicular to the line 5x+2y=12 through the point (-2,3)

muminat

Answer:

The equation of line parallel to given line and passing through points ( - 2 , 3 ) is 5 x + 2 y + 4 = 0

The equation of line perpendicular to given line and passing through points ( - 2 , 3 ) is 2 x - 5 y + 19 = 0

Step-by-step explanation:

Given equation of line as :

5 x + 2 y = 12

or, 2 y = - 5 x + 12

or , y = $\frac{-5}{2}$ x + $\frac{12}{2}$

Or, y = $\frac{-5}{2}$ x + 6

∵ Standard equation of line is give as

y = m x + c

Where m is the slope of line and c is the y-intercept

Now, comparing given line equation with standard eq

So, The slope of the given line = m = $\frac{-5}{2}$

Again,

The other line if passing through the points (- 2 , 3 ) And is parallel to given line

So, for parallel lines condition , the slope of both lines are equal

Let The slope of other line = M

So, M = m = $\frac{-5}{2}$

∴ The equation of line with slope M and passing through points ( -2 , 3) is

y = M x + c

Now , satisfying the points

So, 3 = $\frac{-5}{2}$ × ( - 2 ) + c

or, 3 = $\frac{10}{2}$ + c

Or, 3 = 5 + c

∴ c = 3 - 5 = - 2

c = - 2

So, The equation of line with slope $\frac{-5}{2}$ and passing through points ( -2 , 3)

y = $\frac{-5}{2}$ x - 2

or, 2 y = - 5 x - 4

I.e 5 x + 2 y + 4 = 0

<u>Similarly</u>

The other line if passing through the points (- 2 , 3 ) And is perpendicular to given line

So, for perpendicular lines condition,the products of slope of both lines = - 1

Let The slope of other line = M'

So, M' × m = - 1

Or, M' × $\frac{-5}{2}$ = - 1

Or, M' = $\frac{-1}{\frac{-5}{2}}$

Or, M' = $\frac{2}{5}$

∴ The equation of line with slope M and passing through points ( -2 , 3) is

y = M' x + c'

Now , satisfying the points

So, 3 = $\frac{2}{5}$ × ( - 2 ) + c'

or, 3 = $\frac{- 4}{5}$ + c'

Or, 3 × 5 = - 4 + 5× c'

∴ 5 c' = 15 + 4

or, 5 c' = 19

Or, c' = $\frac{19}{5}$

So, The equation of line with slope $\frac{2}{5}$ and passing through points ( -2 , 3)

y = $\frac{2}{5}$ x + $\frac{19}{5}$

y = $\frac{2 x + 19}{5}$

Or, 5 y = 2 x + 19

Or, 2 x - 5 y + 19 = 0

Hence The equation of line parallel to given line and passing through points ( - 2 , 3 ) is 5 x + 2 y + 4 = 0

And The equation of line perpendicular to given line and passing through points ( - 2 , 3 ) is 2 x - 5 y + 19 = 0

Answer

4 0

3 years ago

I need help with this please

Elena-2011 [213]

Answer:

10(5v-6)

Step-by-step explanation:

7 0

3 years ago