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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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The 10-kg block is held at rest on the smooth inclined plane by the stop block at A. If the10-g bullet is traveling at 300m/swhe
WARRIOR [948]

Answer:

6.8 mm

Explanation:

We are given that

Mass of block,m=10 kg

Mass of bullet,m_b=10 g=10\times 10^{-3} kg

1 kg=1000 g

Total mass of system,M=m+m_a=10+10\times 10^{-3}=10.01kg

Speed of bullet,u=300 m/s

\theta=30^{\circ}

By law of conservation of momentum

m_bucos\theta=Mv

v=\frac{m_bvcos\theta}{M}=\frac{0.01\times 300cos30^{\circ}}{10.01}=0.259m/s

According to law of conservation of energy

Change in kinetic energy of system=Change in potential energy of system

\frac{1}{2}Mv^2-0=Mgh-0

\frac{1}{2}(10.01)(0.259)^2=10.01\times 9.8 h

Where g=9.8 m/s^2

h=\frac{(0.259)^2}{2\times 9.8}=0.0034m

1m=100 cm

h=0.0034\times 100=0.34 cm

Distance traveled by block=d=\frac{h}{sin\theta}=\frac{0.34}{sin30^{\circ}}=0.68 cm=6.8 mm

1cm=10 mm

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