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Yuki888 [10]
3 years ago
7

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.
Physics
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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Why are Mars and Europa the top targets for the study of astrobiology?
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The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde
Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

S_1 S_2=3m

S_1 O=4m

By Pythagoras theorem

                                 S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m

Now

The intensity at O when both speakers are on is given by

I=4I_1 cos^2(\pi \frac{\delta}{\lambda})

Here

  • I is the intensity at O when both speakers are on which is given as 6 W/m^2
  • I1 is the intensity of one speaker on which is 6  W/m^2
  • δ is the Path difference which is given as

                                           \delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m

  • λ is wavelength which is given as

                                             \lambda=\frac{v}{f}

      Here

              v is the speed of sound which is 320 m/s.

              f is the frequency of the sound which is to be calculated.

                                  16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )

where k=0,1,2

for minimum frequency f_1, k=1

                                  {f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz

So the minimum frequency is 702.22 Hz

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Explain why the same side of the moon is always facing Earth.
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What does acceleration mean?
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