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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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Answer:

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5 0
3 years ago
A bowling ball has an initial momentum of +30 kg m/s and hits a stationary bowling pin. After the collision, the bowling ball le
saw5 [17]

Answer:

Momentum of the bowling pin is 17kgm/s

Explanation:

P(initial of ball)=+30 kg m/s

P(initial of pin)=0  because its velocity is 0

P initial = P final

30= P(ball) + P(pin)

30=13+P(pin)

P(final of pin)=+13 kg m/s and P(final of ball)= 30-13=17kgm/s

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3 years ago
A diver springs upward from a board that is 4.40 m above the water. At the instant she contacts the water her speed is 13.5 m/s
Yakvenalex [24]

The diver has the initial velocity, both (a) magnitude is 9.8 m/s and (b) direction is  73.5°.

<h3>What is free falling?</h3>

When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy(P.E)

U =mgh

When the object strikes the ground, all the potential energy converted into kinetic energy.

K.E = 1/2mv²

where v is the speed just before hitting the ground.

A diver springs upward from a board that is 4.40 m above the water. At the instant, she contacts the water her speed is 13.5 m/s and her body makes an angle of 78.1 ° with respect to the horizontal surface of the water.

(a)

From energy conservation principle, initial and final mechanical energy are equal.

1/2mu² + mgh = 1/2mv²

where, u is the initial velocity of the diver.

u = sq rt  (v² - 2gh)

u = sq rt (13.5² - 2x9.81x4.4)

u = 9.798 m/s or 9.8 m/s

Thus, the velocity of the diver is 9.8 m/s.

(b)

The horizontal component of velocity will remain constant.

The horizontal component of acceleration is zero.

Then,

ucosθ = vcosΦ

θ = cos⁻¹ [ (13.5 x cos 78.1)/9.8 ]

θ = 73.5°

Thus, the direction of velocity is  73.5°.

Learn more about free falling.

brainly.com/question/13299152

#SPJ1

5 0
2 years ago
What difficulty will you encounter if you only have data from two recording station?
denpristay [2]

<span>If you have only two data from two recording stations then you will be having a hard time finding the correct location of the epicenter. This is because triangulation method requires 3 recording station. If you have 2 recording station, the 2 circles will intersect at 2 points giving you 2 locations that could possibly be the epicenter.</span>

6 0
3 years ago
Read 2 more answers
Power selection feature for resistors to become water modules 10 liters of water at 25°C to đến
Mila [183]

Answer:

P = 2439.5 W = 2.439 KW

Explanation:

First, we will find the mass of the water:

Mass = (Density)(Volume)

Mass = m = (1 kg/L)(10 L)

m = 10 kg

Now, we will find the energy required to heat the water between given temperature limits:

E = mCΔT

where,

E = energy = ?

C = specific heat capacity of water = 4182 J/kg.°C

ΔT = change in temperature = 95°C - 25°C = 70°C

Therefore,

E = (10 kg)(4182 J/kg.°C)(70°C)

E = 2.927 x 10⁶ J

Now, the power required will be:

Power = P = \frac{E}{t}

where,

t = time = (20 min)(60 s/1 min) = 1200 s

Therefore,

P = \frac{2.927\ x\ 10^6\ J}{1200\ s}

<u>P = 2439.5 W = 2.439 KW</u>

7 0
3 years ago
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