How much salt is in the tank after 30 minutes?

How much salt is in the tank after 30 minutes?

After 30 minutes, there will be 1.448 kg of salt in the tank.

How do you calculate differential equalizer?

Steps

1. Substitute y = uv, and.
2. Factor the parts involving v.
3. Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step)
4. Solve using separation of variables to find u.
5. Substitute u back into the equation we got at step 2.
6. Solve that to find v.

How much salt is in the tank when the tank is full?

The tank will be full when t=15 minutes. The final answer is therefore: Q(15)=225+450+7530=75030=25 pounds of salt.

Is differential equation difficult?

differential equations in general are extremely difficult to solve. thats why first courses focus on the only easy cases, exact equations, especially first order, and linear constant coefficient case. the constant coefficient case is the easiest becaUSE THERE THEY BEhave almost exactly like algebraic equations.

What can differential equations model?

Differential equation models are used in many fields of applied physical science to describe the dynamic aspects of systems. The typical dynamic variable is time, and if it is the only dynamic variable, the analysis will be based on an ordinary differential equation (ODE) model.

How much salt is in a saltwater tank?

A tank contains 80 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drains from the tank at the rate 7 L/min.

How much salt does it take to fill a well tank?

Water containing 0.5 lbs of salt per gallon then enters the tank at a rate of 4 gal/min while the well mixed solution exits the tank at a rate of 1 gal/min. a) How long before the tank begins to overflow?

What is the difference between Tank 1 and Tank 2?

Tank 1 contains 800 liters of water initially containing 20 grams of salt dissolved in it and tank 2 contains 1000 liters of water and initially has 80 grams of salt dissolved in it. Salt water with a concentration of ½ gram/liter of salt enters tank 1 at a rate of 4 liters/hour.

What is the equation for salt in a 500 gallon tank?

= 2 − D / 10 lb/min with initial condition D(0) = 0. The solution to this equation is D(t) = 20(1 − e − t / 10) where t is in (0, ∞) A tank with a capacity of 500 gallons originally contains 200 gallons of water with 50 lbs of salt in solution.