Please help me with these two problems, there pretty easy, I'm just over complicating things to the point I don't know what to do now. Thanks for helping!
A radioactive substance is decaying so that the number of grams present after "t" days is given by the function A(t)=2000e^ -.02t
(the above function is A(t)=2000e to the power of negative .02t ... Just wanted to clear that up)
a. Find the amount of the substance, to the nearest tenth of a gram, present after 30 days.
b. Find the half-life of the substance
2006-11-17 17:32:32 · 4 answers · asked by Shawn 1 in Science & Mathematics ➔ Mathematics
****** I think both of you made mistakes in your problems, if you check the original problem that I listed, the value next to "T" is not .2 it's .02
so that would make the answer of the first problem 1097.6 right? Now I don't know which B part of the problem is correct from you guys. Can some one clear this up?******
2006-11-18 06:19:13 · update #1
EDIT:
a)
Just plug in 30 for 10 in the equation:
A(30) = 2000·e^-(0.02*30) = 1097.6 g (that's to the nearest 10th)
b)
You want to find the time it takes for the amount of substance to decrease to half its original value.
Original value (i.e. @ t=0):
A(0) = 2000·e^-0 = 2000 g
So you want to find t when A=1000 g, so solve for t:
A = 2000·e^-0.02t
e^-0.02t = A / 2000
Now take the natural log (ln) of both sides:
-0.02t = ln(A / 2000)
t = (-1 / 0.2)*ln(A / 2000)
= -50*ln(1000 / 2000)
= -50*ln(0.5)
= 34.66 days
[my part (b) is correct]
2006-11-17 17:40:27 · answer #1 · answered by Anonymous · 0⤊ 0⤋
a) replace t by 30
you obtain A(30) = 2000 e^(-0.2*30) = 2000 e^(-6)= 2000*0.0024 =5
I write 5 since you must give an integer!
b) the half live is the time were half atoms have decayed
so time after which, A(1/2) = 1000
1000 = 2000 e^(-0.21* t(1/2)) divide by 1000
1/2 = e^(-0.21* t(1/2)) 2 = e^(0.21* t(1/2)) go to the logs
ln 2 = 0.21* t (1/2) =3.3 days
remark t(1/2) is called the period
2006-11-17 21:30:13 · answer #2 · answered by maussy 7 · 0⤊ 0⤋
assume that today there is 10 grams of substance, while 1000years ago there was 100 grams of it. If there is 15grams of substance today, how much will there be 600 years from now?
2015-03-23 05:59:23 · answer #3 · answered by cherietha 1 · 0⤊ 0⤋
fixing the differential equation A(t) = C e^ ( -3t /250) a million/2 existence ability the time it reduces to a million/2 of its unique fee this is C then C/2 =C e^( -3t / 250 ) fixing this t = (250 /3) ln(2) = fifty seven.seventy six years
2016-10-15 16:58:47 · answer #4 · answered by Anonymous · 0⤊ 0⤋