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alekssr [168]
3 years ago
6

Interstellar dust causes stars to appear bluer than they actually are

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
4 0
Im not sure what your question is but I will just elaborate. Interstellar dust absorbs more red light than blue light, making stars appear bluer than their true color.
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The answer is C.

The Kinetic energy which was exerted and experience pulling the string of a bow is kept as a potential energy at the end of the arrow in contact with the string. Once release from aim at stationary position the potential energy is again transformed.
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Do you think a hurricane can be considered to be a heat machine? Why?
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Brainliest if correct Question 1 of 10
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What humidity reading would be the driest?
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3 years ago
A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact
ArbitrLikvidat [17]

Answer:

  • 0.09 % of the original radioactive nucllde its left after 10 half-lives
  • It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

N ( t) \ = \ N_0 \ e^{ \ -  \frac{t}{\tau}},

where N(t) its quantity of material at time t, N_0 its the initial quantity of material and \tau its the mean lifetime of the radioactive element.

The half-life t_{\frac{1}{2}} its the time at which the quantity of material its the half of the initial value, so, we can find:

N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

so:

\ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}

-  \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )

t_{\frac{1}{2}}\ = \tau ln( 2 )

So, after 10 half-lives, we got:

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  t_{\frac{1}{2}}}{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  \tau \ ln( 2 ) }{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  10 \  \ ln( 2 ) }

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

10 \ * \ 24,110 \ years \ = \ 241,100 \ years

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

7 0
3 years ago
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