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3 years ago

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Nuclear waste disposal is one of the largest issues with nuclear power. Cesium-137 is one of the high level waste products in an

uranium nuclear fission reaction. Safely disposing of Cs-137 is extremely important due both to this isotope's long half-life, 30.07 years, and it ability to substitute for potassium in biological reactions. A 5.2 mg sample of Cs-137 requires half-lives to decay to 0.65 mg. After 150 years, of Cs-137 remains. After 6 half-lives, of Cs-137 remains. After years, 0.0102 mg of Cs-137 remains. Finally, after approximately 601 years - half-lives, only of the Cs-137 remains.

1 answer:

vodomira [7]3 years ago

7 0

Answer:

A sample of 5.2 mg decays to .65 mg or to 1/8 of its original amount.

1/8 = 1/2 * 1/2 * 1/2 or 3 half-lives.

3 * 30.07 = 90 yrs for 5.2 mg to decay to .65 mg

You can get these other numbers similarly:

5.2 / .0102 = 510 requires about 9 half-lives which is 30 * 9 = 270 yrs

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Where

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$\theta =$ Angular displacement

Our values are given as

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$\omega_f = 47.12rad/s$

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$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

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