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.
The correct answer is D.
A nucleon<span> is one of either of the two types of subatomic particles (neutrons and protons) which are located in the nucleus of atoms.
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The total number of nucleon in the nucleus of an atom gives you an idea about the mass of that atom. In fact, one may refer mass number as nucleon number.
Simply put, nucleons are the particles that make nucleus of an atom and are held up together inside the nucleus due to nuclear force.
Answer:
Explanation:
kinetic energy required = 1.80 MeV
= 1.8 x 10⁶ x 1.6 x 10⁻¹⁹ J
= 2.88 x 10⁻¹³ J
If v be the velocity of proton
1/2 x mass of proton x v² = 2.88 x 10⁻¹³
= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³
v² = 3.45 x 10¹⁴
v = 1.86 x 10⁷ m /s
If V be the potential difference required
V x e = kinetic energy . where e is charge on proton .
V x 1.6 x 10⁻¹⁹ = 2.88 x 10⁻¹³
V = 1.8 x 10⁶ volt .
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13 g —> 0.013 kg
KE = 1/2(m)(v)^2
KE = 1/2(0.013)(8.5)^2
KE = 0.47 J
