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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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2 a A pile of 60 sheets of paper is 6 mm high. Calculate the average thickness of a sheet of the paper.
fiasKO [112]

The average thickness of a sheet of the paper is 0.1 mm.

The number of ice blocks that can be stored in the freezer is 80 blocks of ice.

<h3>Average thickness of a sheet of the paper</h3>

The average thickness of a sheet of the paper is calculated as follows;

average thickness = 6 mm/60 sheets = 0.1 mm /sheet

Thus, the average thickness of a sheet of the paper is 0.1 mm.

<h3>Volume of each block of ice</h3>

Volume = 10 cm x 10 cm x 4 cm

Volume = 400 cm³

<h3>Volume of the freezer</h3>

Volume = 40 cm x 40 cm x 20 cm = 32,000 cm³

<h3>Number of ice blocks that can be stored</h3>

n = 32,000 cm³/400 cm³

n = 80 blocks of ice

Thus, the number of ice blocks that can be stored in the freezer is 80 blocks of ice.

Learn more about average thickness here: brainly.com/question/24268651

#SPJ1

6 0
1 year ago
The total distance traveled divided by the time it takes to travel the distance is
Westkost [7]

"The total distance traveled divided by the time it takes to travel the distance"

That's actually a pretty good definition of average speed. <em>(A)</em>

5 0
3 years ago
How fast is Barry Allen ( The Flash) in season 5 of the Flash?
Burka [1]

Answer:  The Flash, Allen's top speed is Mach 3.3, or 2,532 miles per hour.

Explanation:

8 0
2 years ago
A train at rest emits a sound at a frequency of 1057 Hz. An observer in a car travels away from the sound source at a speed of 2
il63 [147K]

Answer:

993.52 Hz

Explanation:

The frequency of sound emitted by the stationery train is 1057 Hz.

The car travels away from the train at 20.6 m/s.

The frequency the observer hears is given by the formula:

f_o = \frac{v - v_o}{v}f

where v = velocity of sound = 343 m/s

vo = velocity of observer

f = frequency from source

This phenomenon is known as Doppler's effect.

Therefore:

f_o = \frac{343 - 20.6}{343} * 1057\\ \\f_o = 322.4 / 343 * 1057\\\\f_o = 993.52 Hz

The frequency heard by the observer is 993.52 Hz.

4 0
3 years ago
A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn
Lelechka [254]

Answer:

a. 16 s b. -1.866 kJ

Explanation:

a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.

We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².

Since the turntable stops at ω = 0, the time it takes to stop is gotten from

ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.

So it takes the turntable 16 s to stop.

b. The workdone by the turntable to stop W equals its rotational kinetic energy change.

So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ

3 0
2 years ago
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