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4 years ago

State the methods of nuclear fission waste disposal and the suitability of each method.

1 answer:

5 0

Most low-level radioactive waste (LLW) is typically sent to land-based disposal immediately following its packaging for long-term management. This means that for the majority (~90% by volume) of all of the waste types produced by nuclear technologies, a satisfactory disposal means has been developed and is being implemented around the world.


Radioactive wastes are stored so as to avoid any chance of radiation exposure to people, or any pollution.The radioactivity of the wastes decays with time, providing a strong incentive to store high-level waste for about 50 years before disposal.Disposal of low-level waste is straightforward and can be undertaken safely almost anywhere.Storage of used fuel is normally under water for at least five years and then often in dry storage.Deep geological disposal is widely agreed to be the best solution for final disposal of the most radioactive waste produced.

I suggest this site on this subject http://www.world-nuclear.org/information-library/nuclear-fuel-cycle/nuclear-wastes/storage-and-dispo...

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What kind of energy is stored when you squeeze a mattress?

yuradex [85]

Elastic Energy is stored in it.

4 0

3 years ago

Read 2 more answers

Two loudspeakers placed 8.0 m apart are driven in phase by an audio oscillator whose frequency range is 2.2 kHz to 2.9 kHz. A po

My name is Ann [436]

Answer:

The answer to the question is 2.2khz

Explanation:

Let z₁ = 5.4m

Let z₂ = 4.6m

The path difference Δz = z₁-z₂ = 5.4 - 4.6 = 0.8m

For the interference= Δz λ, 2λ, 3λ......

The wavelength λ = 0.8m

The speed of sound v = 344m/s

The frequency f = v/λ = 344/0.8 = 430hz

Now,

f₁ =f, f₂= 2f, f₃ = 3f, f₄= 4f, f₅ =5f which is,

f₁ =f = 430Hz, f₂=2f =860Hz, f₃ =3f =1290Hz f₄ =4f =1720Hz and f₅=5f =2150Hz

f5 = 2120Hz = 2.200Hz

we will convert to two significant figures =2.2kHz

8 0

3 years ago

You are sitting 3 m away from you friend who is watching a cartoon on his phone. How will the sound itensity change if your frie

Zinaida [17]

Answer:

Decreases by $3.6 \times 10^{-3}$ times

Explanation:

The intensity of a sound is defined as the energy of the sound that is flowing in an unit time through the unit area which is in the direction that is perpendicular to the direction of the sound waves movement.

The intensity of energy is described by the inverse square law. It states that the intensity varies inversely with the distance square of the distance.

In other words, the sound intensity decreases as inversely proportional to the squared of the distance. i.e. $\frac{1}{r^2}$

In the context when the distance was 3 m, the intensity of the sound was = $\frac{1}{9}$

But when the distance became 6 cm or 0.06 m, the sound intensity decreases by = $\frac{1}{0.06^2}$

= $3.6 \times 10^{-3}$ times

3 0

3 years ago

PLEASE HELP (WILL GIVE BRAINLIEST TO BEST ANSWER!!!!!)

Luden [163]

Answer:

1) a radio are uses by astronomy

2) 6 bilion waves

3) expert vertified

Explanation:

1) in contrast to an "ordinary" telescope, which receives visible light, a radio telescope "sees" radio waves emitted by radio sources, typically by means of a large parabolic ("dish") antenna, or arrays of them.and Radio telescopes are also the primary means to track space probes, and are used in the SETI project. so must been

radio are almostly

ceiver by astronomy

2) Radio waves have the longest wavelengths in the electromagnetic spectrum.

3)Expert Verified

Radio telescopes are telescopes that are specially designed for observation of long light wavelengths

CARRY ON ✨

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3 years ago

In this problem, you will calculate the location of the center of mass for the Earth-Moon system, and then you will calculate th

Radda [10]

Answer:

a) Option D is correct.

The center of mass between the Eartg and the moon is inside the Earth.

Explanation:

Given,

Mass of the moon = (7.35×10²²) kg

Mass of the Earth = (6.00×10²⁴) kg

Mass of the Sun = (2.00×10³⁰) kg

Distance between the Earth and the moon = (3.80×10⁵) km

Distance between the Earth and the Sun = (1.50×10⁸) km

With the assumption that all.of the bodies being considered are on the same straight line on the x-axis,

Note that Centre of mass is given as

C.M = (Σmx)/(Σm)

For the Earth-moon system, let the earth be x=0, then the moon is at x = (3.80 × 10 5) km away.

C.M = (Σmx)/(Σm)

Σmx = (6.00×10²⁴)) × (0) + (7.35×10²²) × (3.80×10⁵) = (2.793 × 10²⁸) kg.km

Σm = (6.00×10²⁴) + (7.35×10²²) = (6.0735 × 10²⁴) kg

CM = (2.793 × 10²⁸) ÷ (6.0735 × 10²⁴)

CM = (4.60 × 10³) km = 4600 km

This means the centre of mass is 4600 km from the Earth.

The Earth's radius = 6378 km

Hence, the centre if mass is inside the Earth.

Hope this Helps!!!

8 0

3 years ago