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melbel
Group Title
What is the rate of decay of awesome10 which has a halflife of 650 days?
 one year ago
 one year ago
melbel Group Title
What is the rate of decay of awesome10 which has a halflife of 650 days?
 one year ago
 one year ago

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melbel Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

Spectrum Group TitleBest ResponseYou've already chosen the best response.0
if it has a half life of 650 time units it means every 650 time units 1/2 of the samle dissapears.
 one year ago

melbel Group TitleBest ResponseYou've already chosen the best response.0
Yep. :) How would I find the rate of decay, though?
 one year ago

Spectrum Group TitleBest ResponseYou've already chosen the best response.0
halflife= ln2/k that is the half life equation
 one year ago

melbel Group TitleBest ResponseYou've already chosen the best response.0
I have a formula \[N=N _{0}e ^{kt}\] How/where would I plug everything in?
 one year ago

melbel Group TitleBest ResponseYou've already chosen the best response.0
I'm not sure I understand @Spectrum
 one year ago

prettyboy20202 Group TitleBest ResponseYou've already chosen the best response.0
I know spectrum
 one year ago

prettyboy20202 Group TitleBest ResponseYou've already chosen the best response.0
i gave him all right answers
 one year ago

prettyboy20202 Group TitleBest ResponseYou've already chosen the best response.0
for his teset he did good to
 one year ago

ktnguyen1 Group TitleBest ResponseYou've already chosen the best response.0
I think they are asking for k you got N=1/2No t=650 so you plug all numbers in and find k
 one year ago

melbel Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

melbel Group TitleBest ResponseYou've already chosen the best response.0
\[N=\frac{ 1 }{ 2 }N _{0}^{650k}\] Okay, so what do I do next?
 one year ago

melbel Group TitleBest ResponseYou've already chosen the best response.0
Would I put some test number in for No? Like 100 or something?
 one year ago

ktnguyen1 Group TitleBest ResponseYou've already chosen the best response.0
no,dw:1354966748637:dw you plug 1/2No for Nthen you can simplify for No
 one year ago

ktnguyen1 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354967053153:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354952485674:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354952600983:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354952743876:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354953110349:dw
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
I think now it makes sense. Doesn't it?
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
KIS means keep it simple.
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
Derp still need help understanding =/
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
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 \[A=ce^{kt}\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
now we're given two values to find what c is you need to know that \[N(0)=N_0\] \[N_0=ce^0=c\] \[N=N_0e^{kt}\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
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 \[N_0\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{.5N_0}{N_0}=e^{650k}\] the initial N's cancel so you get \[.5=e^{650k}\] to get rid of e take the ln of both sides \[ln(.5)=650k\] \[\frac{ln(.5)}{650}=k\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
your rate of decay is \[\frac{ln(.5)}{650}=k\] your equation will be \[N=N_0e^{\frac{ln(.5)t}{650}}\]
 one year ago

ktnguyen1 Group TitleBest ResponseYou've already chosen the best response.0
@Outkast3r09 : same thing I did. That is why I was confused with @mahmit2012 had done
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{dP}{dt}=kP\] \[\frac{dP}{dt}kP=0\] \[e^{\int{kdt}}=e^{kt+c}=ce^{kt}\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
you integrate the left side you'll just get that and the right side will be P
 one year ago
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