Answer:
The speed in still water is 5 miles per hour.
Step-by-step explanation:
distance with current = 90 miles
distance against current = 10 miles
speed in still water = s
speed of current = 4 mph
speed with current = s + 4
speed against current = s - 4
time = t
speed = distance/time
distance = speed * time
With current:
90 = (s + 4) * t
Against the current:
10 = (s - 4) * t
We have a system of equations:
90 = (s + 4) * t
10 = (s - 4) * t
90 = ts + 4t
10 = ts - 4t
Subtract the second equation from the first equation.
80 = 8t
10 = t
t = 10
10 = t(s - 4)
10 = 10(s - 4)
1 = s - 4
s = 5
Answer: The speed in still water is 5 miles per hour.
3 times 7= 21 times three =63
Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c

when t = 0, Q = 200 L × 1 g/L = 200 g

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min