Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 L of a dyesolution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water owing in at the rate of 2 L/min, the well-stirred solution owing out at the same rate.
a. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value?

Time,t that it will elapse before the concerntration of the dye in the tank reaches1% of its original value is 460.52 seconds

Step-by-step explanation:

Let Q(t) = amount of dye for all time,t.

Let Q'= rate in - rate out.

But Q/200 = concerntration of dye

Therefore rate out= Q/200 ×2.

Q'/Q = - 1/100

Dividing by Q gives

Ln/Q/ + c = -1/100t + c1

Integrating both sides with respect to t

Ln/Q/ = -1/100t + c2.

c1 and c1 are just another constants

Q= c3e^-t/100

Exponentiating both sides will cause the absolute values to get absorbed into the constants giving rise to a new constant c3

200= c3-0/100

c3= 200

2= 200e^-t/100

Ln1/100= - t/100

Ln0.01 = - t/ 100

Cross multiply

t= Ln0.01 × 100

t = 4.6052 ×100

t = 460.52seconds

answer: idk

step-by-step explanation:

54 km / hours

step-by-step explanation:

we need to convert seconds to hours

60 seconds = 1 minutes

60 minutes = 1hour

15 m         60 second   60 minutes

  * *         = 54000 meters/ hour

second     1 minute             1 hour

now we need to convert meters to km

1000 m = 1 km

54000 m       1 km

*   = 54 km / hours

hour               1000m