What is non-point source pollution and why is this kind of pollution hard to monitor and control?
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Answer:
b > 66.41 kg/s
Explanation:
The spring force F = -kx, where k = spring constant, the damping force f = -bv. The net force F' = F + f
F + f = ma
-kx - bv = ma
-kx -bdx/dt = md²x/dt².
Re-arranging the equation, we have
So, md²x/dt² + bdx/dt + kx = 0
Dividing through by m, we have
d²x/dt² + (b/m)dx/dt + (k/m)x = 0
This is a second-order differential equation. The characteristic equation is thus,
D² + (b/m)D + (k/m) = 0
Using the quadratic formula, we find D.
For an overdamped system,
Now, k = F/x. Since the weight of the object causes the spring to stretch a distance of 0.5 m, k = mg/x where m = mass of object = 0.75 kg, g = 9.8 m/s² and x = x₀ =0.5 m.
Substituting k = mg/x into the inequality for b, we have
b > 2√{(mg/x₀)m}
b > 2√{(m²g/x₀)}
b > 2m√{g/x₀)}
b > 2 × 0.75 kg√{9.8 m/s²/0.5 m)}
b > 1.5 kg√{19.6/s²)}
b > 1.5 kg × 4.427/s
b > 66.41 kg/s
The mechanical energy of the river water per unit mass is 887.4 J/kg
The power generated is 452.574 MW.
Given:
Average velocity (v) = 3 m/s
Rate = 510 m³/s
Height (h) = 90 m
We know, that mechanical energy is the sum of potential energy and kinetic energy.
So,
E = ×m×v² + m×g×h
and energy per mass unit is
E/m = ×v² + g×h
E/m = ×3² + 9.81×90
E/m = 887.4 J/kg
So, mechanical energy per unit mass is 887.4 J/kg.
Power generated is expressed as;
Power generated = energy per unit mass ×rate×density
Density of water = 1000 kg/m³
Power generated = 887.4× 510× 1000
Power generated = 452574000 W
So, the power generated is 452.574 MW.
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