anonymous
  • anonymous
What is the rate of decay of awesome-10 which has a half-life of 650 days?
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
I can't figure out how to solve these rate of decay questions, so I just made up my own as a way to help me learn this. How would I solve this?
anonymous
  • anonymous
if it has a half life of 650 time units it means every 650 time units 1/2 of the samle dissapears.
anonymous
  • anonymous
Yep. :) How would I find the rate of decay, though?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
halflife= -ln2/k that is the half life equation
anonymous
  • anonymous
I have a formula \[N=N _{0}e ^{kt}\] How/where would I plug everything in?
anonymous
  • anonymous
I'm not sure I understand @Spectrum
anonymous
  • anonymous
I know spectrum
anonymous
  • anonymous
i gave him all right answers
anonymous
  • anonymous
for his teset he did good to
anonymous
  • anonymous
I think they are asking for k you got N=1/2No t=650 so you plug all numbers in and find k
anonymous
  • anonymous
I mean, I don't understand what he meant. I'm having a lot of trouble with logarithms, so I'm not quick on my feet. :P
anonymous
  • anonymous
\[N=\frac{ 1 }{ 2 }N _{0}^{650k}\] Okay, so what do I do next?
anonymous
  • anonymous
Would I put some test number in for No? Like 100 or something?
anonymous
  • anonymous
no,|dw:1354966748637:dw| you plug 1/2No for Nthen you can simplify for No
anonymous
  • anonymous
|dw:1354967053153:dw|
anonymous
  • anonymous
|dw:1354952485674:dw|
anonymous
  • anonymous
|dw:1354952600983:dw|
anonymous
  • anonymous
|dw:1354952743876:dw|
anonymous
  • anonymous
|dw:1354953110349:dw|
anonymous
  • anonymous
I think now it makes sense. Doesn't it?
anonymous
  • anonymous
KIS means keep it simple.
anonymous
  • anonymous
Derp still need help understanding =/
anonymous
  • anonymous
The easiest way I believe is to derive it from calculus (if you've taken Diff EQ , you should know how to). Doing this you get an equation much like yours only N_0 is denoted c \[A=ce^{kt}\]
anonymous
  • anonymous
now we're given two values to find what c is you need to know that \[N(0)=N_0\] \[N_0=ce^0=c\] \[N=N_0e^{kt}\]
anonymous
  • anonymous
same as your equation now we know that after 650 days , we will have half of the original amount soo \[N(650 days)=.5N_0\] \[.5N_0=N_0e^{650k}\] solve for k, divide by \[N_0\]
anonymous
  • anonymous
\[\frac{.5N_0}{N_0}=e^{650k}\] the initial N's cancel so you get \[.5=e^{650k}\] to get rid of e take the ln of both sides \[ln(.5)=650k\] \[\frac{ln(.5)}{650}=k\]
anonymous
  • anonymous
your rate of decay is \[\frac{ln(.5)}{650}=k\] your equation will be \[N=N_0e^{\frac{ln(.5)t}{650}}\]
anonymous
  • anonymous
@Outkast3r09 : same thing I did. That is why I was confused with @mahmit2012 had done
anonymous
  • anonymous
\[\frac{dP}{dt}=kP\] \[\frac{dP}{dt}-kP=0\] \[e^{\int{kdt}}=e^{kt+c}=ce^{kt}\]
anonymous
  • anonymous
you integrate the left side you'll just get that and the right side will be P

Looking for something else?

Not the answer you are looking for? Search for more explanations.