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4 years ago

Acid mine drainage is:

Physics

1 answer:

Dahasolnce [82]4 years ago

8 0

Answer:

Acid mine drainage is dissolved toxic materials wash from mines into nearby lakes and streams.

Explanation:

Acid mine drainage is the flow of acidic water with pH typically between 2 and 4, and high concentrations of other dissolved toxic materials from mines into nearby lakes and streams. It mainly occurs during metal sulfide mining, when the metal sulfide ore such as pyrite (FeS2) is exposed to water and oxygen from air to produce soluble iron and sulfuric acid.

Microorganisms, especially acidophile bacteria like Acidithiobacillus ferrooxidans grow by pyrite oxidation, i.e., oxidizing the Fe²⁺ in pyrite to Fe³⁺, which again react with pyrite and water to produce sulfuric acid. Then the acidic water flows into nearby water sources and reduces the pH value of water in those sources. As a result, heavy metals such as copper, lead, mercury, etc in other mineral ores also get dissolved into the water. The action of acidophile bacteria also increases the rate and degree of acid-mine drainage process.

The acid mine drainage causes water pollution and adversely affect the aquatic plants and animals. It also results in the contamination of drinking water, corrosion of infrastructures such as bridges, etc.

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4. What is the momentum of a 70 kg object traveling at 20 m/s?

Soloha48 [4]

Answer:

1400 units of momentum.

Explanation:

Using the formula p=mv. We can get the momentum using 70*20 =1400 units of momentum

6 0

3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

morpeh [17]

Answer:

63.750KeV

Explanation:

We are given that

Initial velocity of second electron, $u_2=0$

Radius, $r_1=0$

$r_2=2.3 cm=\frac{2.3}{100}=0.023 m$

1 m=100 cm

Magnetic field,B=0.0370 T

We have to determine the energy of the incident electron.

Mass of electron, $m=9.1\times 10^{-31} kg$

Charge on an electron, $q=-1.6\times 10^{-19} C$

Velocity, $v=\frac{Bqr}{m}$

Using the formula

Speed of electron, $v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0$

Speed of second electron, $v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}$

$v_2=1.5\times 10^8 m/s$

Kinetic energy of incident electron= $\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2$

Kinetic energy of incident electron= $0+\frac{1}{2}(9.1\times 10^{-31})(1.5\times 10^8)^2=1.02\times 10^{-14} J$

Kinetic energy of incident electron= $\frac{1.02\times 10^{-14}}{1.6\times 10^{-19}}=63750eV=\frac{63750}{1000}=63.750KeV$

1KeV=1000eV

3 0

3 years ago

What is cell in electricity?

agasfer [191]

Answer:

An electrical cell is a device used to generate electricity, or to make chemical reactions by applying electricity.

7 0

4 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

3 years ago

The velocity of a bob on a simple pendulum at the lowest position is 10.56 m/s. What is the maximum vertical height it is able t

kondaur [170]

um I do not know sorry

Explanation:

ummmmmmjjhfgfffff

7 0

3 years ago