6.P.2.1 Matter: Properties and Changes Notes

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

Alexeev081 [22]

<span>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.
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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.<span>Deep geological disposal is widely agreed to be the best solution for final disposal of the most radioactive waste produced.

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

5 0

3 years ago

A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa

Lesechka [4]

At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
$U_i= \frac{1}{2} ka^2$
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
$K_i =0$
And the total energy of the system is
$E_i = U_i+K= \frac{1}{2}ka^2$

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
$U_f = 0$
while the mass is moving at speed v, and therefore the kinetic energy is
$K_f = \frac{1}{2} mv^2$
And the total energy is
$E_f = U_f + K_f = \frac{1}{2} mv^2$

For the law of conservation of energy, the total energy must be conserved, therefore $E_i = E_f$ . So we can write
$\frac{1}{2} ka^2 = \frac{1}{2}mv^2$
that we can solve to find an expression for v:
$v= \sqrt{ \frac{ka^2}{m} }$

6 0

3 years ago

A photon with an energy E = 2.12 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 96.0 MeV. What

Oksanka [162]

Answer:

The kinetic energy of the anti proton is 147.4 MeV.

Explanation:

Given that,

Energy = 2.12 GeV

Kinetic energy = 96.0 MeV

We need to calculate the kinetic energy of the anti proton

Using formula of energy

$E_{photon}=m_{p}c^2+m_{np}c^2+K.E_{p}+K.E_{np}$

We know that,

$m_{p}c^2=m_{np}c^2$

So, $E_{photon}=2mc^2+K.E_{p}+K.E_{np}$

$K.E_{np}=E_{photon}-(2mc^2+K.E_{p})$

Put the value into the formula

$K.E_{np}=2.12\times10^{9}-2\times938.3\times10^{6}-96\times10^{6}$

$K.E_{np}=147.4\ MeV$

Hence, The kinetic energy of the anti proton is 147.4 MeV.

6 0

3 years ago

2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind

Alex777 [14]

The resultant speed of the plane is (3) 226 m/s

Why?

We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).

So, calculating we have:

$ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}$

$ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}$

Hence, we have that the resultant speed of the plane is (3) 226 m/s

Have a nice day!

5 0

3 years ago

A light source emits light with dominant wavelengths in the range of 650 to 690 nm. what is the principle color of light emitted

dsp73

Answer:

Please see below as the answer is self-explanatory.

Explanation:

The visible range extends roughly from 400 nm (violet) to 700 nm (red).
Below the violet is the ultra-violet spectrum (with higher energy) and above red, we have the infra-red spectrum.
The wavelengths in the range of 650 to 690 nm have red as the dominant color.

6 0

2 years ago