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.
Answer: Transverse waves have motion perpendicular to velocity, while longitudinal waves have motion parallel to velocity.
Explanation:
Transverse waves are characterized by the fact that the particles of the medium in which they propagate move transversely to the direction of propagation of the wave.
In other words,<u> its displacement is perpendicular to the direction of propagation of the wave</u>, being a good example the circular waves in the water.
On the other hand, Longitudinal waves are characterized by the fact that <u>the oscillation of the particles in the medium is parallel to the direction of propagation of the wave.</u> A good example of this is the sound wave.
Answer:
V = 10.88 m/s
Explanation:
V_i =initial velocity = 0m/s
a= acceleration= gsinθ-
cosθ
putting values we get
a= 9.8sin25-0.2cos25= 2.4 m/s^2
v_f= final velocity and d= displacement along the inclined plane = 10.4 m
using the equation
v_f= 7.04 m/s
let the speed just before she lands be "V"
using conservation of energy
KE + PE at the edge of cliff = KE at bottom of cliff
(0.5) m V_f^2 + mgh = (0.5) m V^2
V^2 = V_f^2 + 2gh
V^2 = 7.04^2 + 2 x 9.8 x 3.5
V = 10.88 m/s
Σf = m a
Σf = m v^2 / r
Σf = 52 8^2 / 1.6
Σf = 2080 N
7.625 Newtons
work = force× distance
Newtons is an accepted value for force
so take the total 224 joules and decide by distance 32 meters to find force in Newtons
