Answer:
Let's start by understanding what exactly a scientific question is. A scientific question is a question that may lead to a hypothesis and help us in answering (or figuring out) the reason for some observation. A good scientific question has certain characteristics. It should have some answers (real answers), should be testable.
Here's examples of a few:
Why is that a star?
or
What is that star made of?
Hope this can lead you to the answer you're looking for at least!!
Answer:
The torque about his shoulder is 34.3Nm.
The solution approach assumes that the weight of the boy's arm acts at the center of the boy's arm length 35cm from the shoulder.
Explanation:
The solution to the problem can be found in the attachment below.
Answer:
Higher percentage of runoff water is absorbed by the vegetation whereas 100 % of water runoff on the roofed area.
Explanation:
Higher percentage of runoff water is absorbed by the vegetation and very little amount of runoff water is received on the end of vegetative cover while on the other hand, 100 percent of precipitation water runoff on the surface of roofed or paved area because no water is absorbed by the roof due to its solid surface. The percentage of runoff depends on the factors such as rainfall intensity, slope, soil water storage capacity, and infiltration rate.
Answer:
The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L
Explanation:
Let the initial concentration of the BOD = C₀
Concentration of BOD at any time or point = C
dC/dt = - KC
∫ dC/C = -k ∫ dt
Integrating the left hand side from C₀ to C and the right hand side from 0 to t
In (C/C₀) = -kt + b (b = constant of integration)
At t = 0, C = C₀
In 1 = 0 + b
b = 0
In (C/C₀) = - kt
(C/C₀) = e⁻ᵏᵗ
C = C₀ e⁻ᵏᵗ
C₀ = 75 mg/L
k = 0.05 /day
C = 75 e⁻⁰•⁰⁵ᵗ
So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day
We calculate how many days it takes the river to reach 50 km downstream
Velocity = (displacement/time)
15 = 50/t
t = 50/15 = 3.3333 days
So, we need the C that corresponds to t = 3.3333 days
C = 75 e⁻⁰•⁰⁵ᵗ
0.05 t = 0.05 × 3.333 = 0.167
C = 75 e⁻⁰•¹⁶⁷
C = 63.5 mg/L
