Answer:
The answer is c and the teacher helped me
Explanation:
i had help from the tescger and the assignment is done
Answer:
The average velocity is 0.203 m/s
Explanation:
Given;
initial displacement, x₁ = 20 yards = 18.288 m
final displacement, x₂ = ¹/₃ x 18.288 = 6.096 m
change in time between 5:02 PM and 5:03 PM, Δt = 3 mins - 2 mins = 1 min = 60 s
The average velocity is given by;
V = change in displacement / change in time
V = (x₂ - x₁) / Δt
V = (18.288 - 6.096) / 60
V = 0.203 m/s
Therefore, the average velocity is 0.203 m/s
A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s) with a flow rate of 8.7 m^3/s, 6 mg/L BOD5 and 8.3 mg/L DO. Both streams are at 20°C. After mixing, the river is 3 meters deep and flowing at a velocity of 0.50 m/s. DOsat for this river is 9.0 mg/L. The deoxygenation constant is kd= 0.20 d^-1 and The reaction rate constant k at 20 °C is 0.27 d^-1.
The answer therefore would be the number 0.27 divided by two and then square while getting the square you would make it a binomial.
I wont give the answer but the steps
Your Welcome
A. Email your teacher right away. It would be the safest option.
