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Delicious77 [7]
3 years ago
10

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

Physics
2 answers:
ira [324]3 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]3 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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Answer: 71.16\ Hz

Explanation:

Given

Capacitance C=100\ \mu F

Resistance R=500\ \Omega

Inductance L=50\ mH

In LCR circuit, current is maximum at resonance frequency i.e.

X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}

Insert the values

\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}

Also, frequency is given by

\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}

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2 years ago
An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f
IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}

Similarly, The equation of mption:

x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})

replacing our assumed values:

where v_=0 \ and \ g= 9.81

x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})

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So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

1000= 0.981t-0.981(1-e^{-(10)t}) \ m

By diregarding e^{-(10)t} \ m

1000= 0.981t-0.981

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

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