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nirvana33 [79]
2 years ago
10

A 25 MGD surface water drinking plant has four circular clarifiers (aka sedimentation basin), operated in parallel with each bas

in receiving equal flow, that all have a diameter of 16 meters and are 3 m deep. Assume that all particles entering the clarifiers have the same particle density of 1.2 mg/mL and the average water temperature is 20 oC with an average viscosity of 1 g/(m*s) and an average density of 998.2 kg/m3.
1. What is the diameter of the smallest particles (in mm) that can be removed by these clarifiers?
2. If the each of the four clarifiers is to be rectangular in shape with a length to width ratio of 5:1, what is the minimum width that each clarifier can be (in meters)?
Physics
1 answer:
dangina [55]2 years ago
8 0

Answer:

can y

Explanation:

jj

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n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend
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Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

p_f=p_i\\mv_1+Mv_2=(m+M)v

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m

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