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Vladimir79 [104]
3 years ago
5

Which is a concern about mining for uranium? Heated water could be released into the environment. Dust released in the air could

be radioactive. Supplies could become low during a drought. It is too commonly found, and mines could be too plentiful.
Physics
2 answers:
saw5 [17]3 years ago
3 0

Answer: Dust released in the air could be radioactive.

Explanation:

Uranium is a radioactive element. It decays naturally to attain stability. While mining, the dust displaces into the air causing not only harm to environment but for the workers as well. The released in the air could be radioactive which would be inhaled posing threat for lung cancer.

Thus, the correct answer is dust released in the air could be radioactive is a concern about mining Uranium.

WINSTONCH [101]3 years ago
3 0

Answer:

Dust released in the air could be radioactive.

Explanation:

There is significant access to what are naturally occurring radioactive materials (NORM) for individuals involved with mining. As with many other infection control threats, it is also important to control the risks. In practice, dust is really the primary source of exposure to radiation in an accessible-cut uranium mine and in the mill region.

The dust produced in the extraction is toxic during mining process, that can be harmful to the employees involved in the extraction process. The radioactive material may threaten the workers with a circumstance of lung cancer and respiratory disease.

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How old is a bone that has 12.5% of the original amount radioactive carbon 14 remaining?
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Nuclear decay formula is N(t)=N₀*2^-(t/T), where N(t) is the amount of nuclear material in some moment t, N₀ is the original amount of nuclear material, t is time and T is the half life of the material, in this case carbon 14. In our case N(t)=12.5% of N₀ or N(t)=0.125*N₀, T=5730 years and we need to solve for t:

0.125*N₀=N₀*2^-(t/T), N₀ cancels out and we get:

0.125=2^-(t/T), 

ln(0.125)=ln(2^-(t/T))

ln(0.125)=-(t/T)*ln(2), we divide by ln(2),

ln(0.125)/ln(2)=-t/T, multiply by T,

{ln(0.125)/ln(2)}*T=-t, divide by (-1) and plug in T=5730 years,

{ln(0.125)/[-ln(2)]}*5730=t

t=3*5730=17190 years.

The bone is t= 17190 years old. 

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