急!maths!! - Ans.fyi 參考答案

急!maths!!

2011-02-21 10:20 pm

‎跟Continuous Growth and Decay~THX!

3.The mass of radioactive substance decreases by half after 2.77 years. If the mass of the substance is 25g at present, find the mass after each of the following periods.

(a) 5 years
(b) 25 years ‎4.A principal $P is deposited in a bank at an interest rate r% p.a. compounded continuously. If the amount after 12 years doubles the principal, find the interest rate p.a.7.The mass(in grams) of a radioactive substance at time t years after the first observation is given by M=60e^-0.15t

(a) Find the initial mass of the substance.
(b) What will be the mass after 20 years?
⋯⋯(c) Find the half-life of the substance.(Give your ans. corr. to the nearest years.)[The half-life of a radioactive substance is the time taken for its mass to reduce to half] 8.The decay of a radioactive substance is given by M=M0e^-kt ,where M gram is the mass of the substance after t years, M0 and k are positive constants.

(a)If 30% of the mass of the substance will remain after 100 years,find the value of k.
(b)How long will it take for the mass to 15% of the initial mass?

回答 (1)

pingshek

2011-02-22 9:42 am

✔ 最佳答案

3.
(a)
The mass after 5 years
= 25 x (1/2)^(5/2.77) g
= 7.15 g (3 sig. fig.)

(b)
The mass after 25 years
= 25 x (1/2)^(25/2.77) g
= 0.0480 g (3 sig. fig.)

‎4.
P(1 + r%)^12 = 2P
(1 + r%)^12 = 2
1 + r% = 2^(1/12)
r% = 5.95%
Interest rate p.a. = 5.95%

7.
M = 60e^(-0.15t)

(a)
Initial mass of the substance
= 60e(-0.15*0) g
= 60 g

(b)
The mass after 20 years
= 60e(-0.15*20) g
= 3.00 g (3 sig. fig.)

(c)
Let y years be the half-life.

M = 60e^-0.15t ...... [1]
(1/2)M = 60e^-0.15(t + y) ...... [2]

[2]/[1]:
1/2 = e^-0.15y
ln(1/2) = ln(e^-0.15y)
-ln2 = -0.15y
y = ln2/0.15
y = 5 years (to the nearest years)

8.
M = Moe^-kt

(a)
When t = 0 year, M = Mo
When t = 100 years, M = Moe^-100k = 30%Mo
e^-100k = 30%
ln(e^-100k) = ln(30%)
-100k = ln(30%)
k = [ln(30%)] / (-100)
k = 0.0120 year^-1

(b)
Let y years be the time for the mass to 15% of the initial mass.

When t = 0,
M = Moe^0 = Mo

When t = y,
M = Moe^-0.0120y = 15%Mo
e^-0.0120y = 15%
ln(e^-0.0120y) = ln(15%)
-0.0120y = ln(15%)
y = [ln(15%)] / (-0.0120)
y = 158
Time taken for the mass to 15% of the initial mass = 158 years (to the nearest years)

參考: miraco

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