A tank contains 300 liters of fluid in which 40 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate. Let the A(t) be the number of grams of salt in the tank at time t. Find the rate at which the number of grams of salt in the tank is changing at time t. dA dt = Find the number A(t) of grams of salt in the tank at time t. A(t) = g

Step-by-step explanation:

Solution:-

- The amount of salt in the solution is ( A ) at any time t.

- Pure water enters the tank ( no salt ( A = 0 ) ).

- The volumetric rate of flow in and out of tank is V(flow) = 5 L / min  

- The rate of change of salt in the tank at time ( t ) can be expressed as a first order ordinary differential equation for the salt solution that flows in and out of the tank

- The first order ordinary differential equation is expressed as:

                     = ( salt flow in ) - ( salt flow out )

- Fresh water with zero salt content flows in then ( salt flow in ) = 0

- The concentration of salt within the tank changes with time ( t ).

- The volume of water in the tank remains constant ( steady state conditions ). I.e 5 Liters volume leaves and 5 Liters is added; hence, the total volume of solution in tank remains 300 Liters.

- So at any instant the concentration of salt in the 300 Liter tank is:

- The amount of salt-solution flowing out of the tank per unit time would be:

- The differential equation becomes:

- Separate the variables and integrate both sides:

- Initial conditions: A ( 0 ) = 4 grams. Use the initial conditions to evaluate the constant of integration:

- The solution to the differential equation becomes::

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: