A tank initially contains 100 gallons of brine with 10 lb of salt dissolved in it. Brinecontaining 0.4 lb of salt per gallon ows from an outside source into the tank at a rateof 5 gal/min. The mixture is discharged out of the tank at the same rate of 5 gal/min..Assume that solutions are well-stirred at all times. Required:
Find the amount of salt in the tank at the end of 40 minutes.

35.94 lb

Step-by-step explanation:

The net mass flow rate dm/dt = flow rate in - flow rate out

mass flow rate in = concentration in × volume flow rate in = 0.4 lb/gallon of salt × 5 gal/min = 2 lb/min

Let m(t) be the mass present at time, t.

So, the concentration at time, t = m(t)/volume of tank = m(t)/100

mass flow rate out = concentration out × volume flow rate out = m(t)/100 lb/gal× 5 gal/min = m(t)/20 lb/min

So, dm/dt = 2 lb/min - m(t)/20 lb/min

So, dm/dt = 2 - m(t)/20

dm/dt = [40 - m(t)]/20

separating the variables, we have

dm/[40 - m(t)] = dt/20

integrating, we have

∫dm/[40 - m(t)] = ∫dt/20

-1/-1 ×∫dm/[40 - m(t)] = ∫dt/20

1/-1 ×∫-dm/[40 - m(t)] = ∫dt/20

1/-1 ×㏑[40 - m(t)] = t/20 + C

-㏑[40 - m(t)] = t/20 + C

㏑[40 - m(t)] = -t/20 - C

taking exponents of both sides, we have

when t = 0 m(0) = 10 lb

So

So,

when t = 40 minutes, m(40) =

So, at the end of 40 minutes, the amount of salt in the tank is 35.94 lb

30 miles

step-by-step explanation:

we know that 60 seconds equals 1 minute and 60 minutes 1 hour

44 ft       60 sec     60 min     158400   ft

  *   *   =

1 sec       1 min       1 hour         hour

we know that there are 5280 ft per mile

158400   ft       1 mile         30 miles

  *   =

1   hour               5280 ft     1 hour

so in 1 hour   it can travel 30 miles