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yuradex [85]
3 years ago
8

What is the purpose of the coagulation step in water treatment?

Physics
1 answer:
grigory [225]3 years ago
6 0

Answer:

It is, however, an important primary step in the water treatment process, because coagulation removes many of the particles, such as dissolved organic carbon, that make water difficult to disinfect. Because coagulation removes some of the dissolved substances, less chlorine must be added to disinfect the water.

Explanation:

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7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
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Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

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The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
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