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Delicious77 [7]
4 years ago
10

How was the water filtered to remove debris and living organisms?

Physics
2 answers:
ira [324]4 years ago
5 0

<span>The most common method in removing debris and living organisms is through a series of sedimentation and heat application. The role of sedimentation is to use gravity in separating drinkable water and visible debris. While treating water with heat kills the living organisms.</span>

astra-53 [7]4 years ago
4 0

<u>Step-1: Coagulation</u>

  • Coagulation removes fine particles suspended in water, after removing rough debris by filtering through the screen.
  • Alum and other chemicals are added to water to form small sticky particles called floc, which attracts suspended particles.
  • The combined weight of these particles and alum (floc) becomes heavy enough to sink.

<u>Step-2: Sedimentation</u>

Coagulated particles fall gravitationally through the water in a settling tank and accumulate at the bottom of the tank, cleaning water from much solid debris.

<u>Step-3: Filtration</u>

To remove solid particles that were previously unremoved by sedimentation, water from the sedimentation tank flows through:

  • sand,
  • gravel,
  • coal, and
  • activated charcoal.

<u>Step-4: Disinfection</u>

  • Disinfection methods are applied to filtered water to destroy hazardous microorganisms as living organisms.  
  • A chlorine residue is added to water when leaving the processing plant to help protect treated water from recontamination when moving through the distribution system.
<h3>Further explanation</h3>

Chemical precipitation or coagulation with various aluminum salts (e.g., alum), iron, lime, and other organic or inorganic chemicals is widely used to treat water to remove colloidal particles (turbidness) and microbes.

Water treatment with the addition of coagulants and chemical deposits has been carried out since ancient times.

There are main things to know about the different terms of water purification and filtration because they are not synonyms.

  • Water filtration is a treatment for eliminating debris, and some bacteria as living organisms, through several types of filters like cloth or nets.
  • Water purification is a chemical or UV process (with heat treatment) to deactivate bacteria and other hazardous substances that are not functioning.
<h3>Learn more</h3>
  1. The reason why water is a compound brainly.com/question/4636675
  2. How many carbon atoms are there in a 1.3-carat diamond? brainly.com/question/4235993
  3. Types of fatty acids brainly.com/question/1386856

Keywords: how was the water, filtered, to remove, debris and living organisms, microorganisms, bacteria, microbes, colloidal particles, coagulation, sedimentation, filtration, disinfection, purification

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H-a seated high school student

Explanation:

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Among the choices above, the high school student is the heaviest and has greater mass than the leaf, softball or balloon. This means, <u>it has the greatest inertia.</u>

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When you throw a ball, the work you do to accelerate it equals the kinetic energy the ball gains. If you do twice as much work w
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Answer:

No. Twice as much work will give the ball twice as much kinetic energy. But since KE is proportional to the speed squared, the speed will be sqrt{2} times larger.

Explanation:

The work done on the ball is equal to the kinetic energy gained by the ball:

W=K

So when the work done doubles, the kinetic energy doubles as well:

2W = 2 K

However, the kinetic energy is given by

K=\frac{1}{2}mv^2

where

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v is its speed

We see that the kinetic energy is proportional to the square of the speed, v^2. We can rewrite the last equation as

v=\sqrt{\frac{2K}{m}}

which also means

v=\sqrt{\frac{2W}{m}}

If the work is doubled,

W'=2W

So the new speed is

v'=\sqrt{\frac{2(2W)}{m}}=\sqrt{2}\sqrt{\frac{2W}{m}}=\sqrt{2} v

So, the speed is \sqrt{2} times larger.

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The Sears Tower is nearly 400 m high. How long would it take a steel ball to reach the ground if dropped on the top? What will b
kipiarov [429]

Answers:

a) 9.035 s

b) -88.543 m/s

Explanation:

The described situation is related to vertical motion (especifically free fall) and the equations that will be useful are:

y=y_{o}+V_{o}t+\frac{1}{2}gt^{2} (1)  

V=V_{o}+gt (2)  

Where:  

y=0 is the final height of the steel ball

y_{o}=400 m is the initial height of the steel ball

V_{o}=0 is the initial velocity of the steel ball (it was dropped)

V is the final velocity of the steel ball

t is the time it takes to the steel ball to reach the ground

g=-9.8 m/s^{2} is the acceleration due to gravity

<u>Knowing this, let's begin with the answers:</u>

<h2>a) Time it takes the steel ball to reach the ground</h2>

We will use equation (1) with the conditions listed above:

0=y_{o}+\frac{1}{2}gt^{2} (3)  

Isolating t:

t=\sqrt{\frac{-2y_{o}}{g}} (4)  

t=\sqrt{\frac{-2(400 m)}{-9.8 m/s^{2}}} (5)  

t=9.035 s (6)  

<h2>b) Final velocity of the steel ball</h2>

We will use equation (2) with the conditions explained above and the calculaated time:

V=gt (7)  

V=(-9.8 m/s^{2})(9.035 s) (8)  

V=-88.543 m/s (9)  The negative sign indicates the direction of the velocity is downwards

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