Why must mine tailings be stored and disposed of carefully?

1 answer:

7 0

After thorough researching, the mine tailings must be stored and disposed of carefully because they have lots of chemical and various toxic materials. They can also leach to the aquifers. The correct answer to the following given statement above is they have chemicals which are dangerous.

You might be interested in

Ferromagnetic materials-

QveST [7]

Answer:

materials which exhibit a spontaneous net magnetization at the atomic level, even in the absence of an external magnetic field.

Explanation:

When a material is placed within a magnetic field, the magnetic forces of the material's electrons will be affected. This effect is known as Faraday's Law of Magnetic Induction. However, materials can react quite differently to the presence of an external magnetic field. This reaction is dependent on a number of factors, such as the atomic and molecular structure of the material, and the net magnetic field associated with the atoms. The magnetic moments associated with atoms have three origins. These are the electron motion, the change in motion caused by an external magnetic field, and the spin of the electrons.

5 0

3 years ago

Please help me guys never mind the calculations

vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} \\ \frac{1}{R} = \frac{3 + 2}{6} = \frac{5}{6} \\ R = \frac{6}{5} = 1.2 \: \: ohm$

5.2) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{8} + \frac{1}{10} \\ \frac{1}{R} = \frac{5 + 4}{40} = \frac{9}{40} \\ R = \frac{40}{9} = 4.4 \: \: ohm$

I hope I helped you^_^

7 0

3 years ago

When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th

Veseljchak [2.6K]

Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant

The work done in stretching the rubber band from x = 0 to x = L is
$W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}$

Answer: $\frac{aL^{2}}{2}$

4 0

4 years ago

car was moving in a straight road of length 320 km it covered 240 km with an average velocity 75 km/hr then it ran out of fuel a

Stella [2.4K]

The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters: