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2 years ago

What is a common environmental problem caused by mining

2 answers:

7 0

Erosion, formation of sinkholes, loss of biodiversity, and contamination of soil, groundwater and surface water by chemicals from mining processes.

Anarel [89]2 years ago

3 0

Answer:

Among the main impacts that mining can cause to the environment are the destruction of the earth's crust, the contamination of the waters, the impact on the flora and fauna of the environment close to mining and the negative effects on human health of nearby populations to the mine, as well as deforestation, earthworks, chemical contamination and other substances used to extract the ore.

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A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the

Gennadij [26K]

Answer:

Let I and j be the unit vector along x and y axis respectively.

Electric field at origin is given by

E= kq1/r1^2 i + kq2/r2^2j

= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)

= (2.88i + 1.44j)*10^-3 N/C

Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3

F= (1.382 i + 0.691 j) *10^-21

Goodluck

Explanation:

4 0

1 year ago

A rightward force of 12.0 N is applied to a 2.0-kg object to accelerate it across a horizontal

ad-work [718]

Answer:

below

Explanation:

Net accelerating force becomes 12-8 = 4 N

F = ma

4 = 2 * a

a = 2 m/s^2

8 0

10 months ago

State the name of the process in each of the following changes of state:

polet [3.4K]

Answer:

A. Melting

B. Freezing

C. Condensing

D. Boiling

Explanation:

4 0

3 years ago

A car is traveling north at 17.7 m/s . After 6 it’s velocity is 141 in the same direction. Find the magnitude and direction of t

Furkat [3]

By equation of motion we have v = u + at

Where u = Initial velocity, v = final velocity, t = time taken and a = acceleration

Here v = 141 m/s, u = 17.7 m/s and t = 6 s

On substitution we will get

141 = 17.7+ 6a

So, a = (141-17.7)/6 = 20. 55 m/ $s^{2}$

Aceeleration = 20. 55 m/ $s^{2}$ along north direction.

3 0

3 years ago

A system consists of electrons and protons only. It contains 150 electrons and has a total charge of +22e. What is the mass of t

liubo4ka [24]

According to the statements the number of electrons is 150, then

e = 150

But there is a positive charge of +22e, then the number of protons would be

p = 150+172

If the mass of the electrons is

$m_e = 9.11*10^{-31} kg$

And the mass of the protong is

$m_p = 1.673*10^{-27}kg$

We have that the total mass of the system would be

$m = e*m_e +pm_p$

$m = 150 * (9.11*10^{-31})+170(1.673*10^{-27})$

$m = 2.84547*10^{-25} kg$

5 0

3 years ago