數學的單元一問題 - Ans.fyi 參考答案

數學的單元一問題

Bhine

2010-10-24 8:16 am

在一個化學反應中,反應物的濃度在5分鐘內由0.037mol/L減至0.012mol/L.假設濃度以固定的率減少,8分鐘內濃度會減少至多少??準確至3位小數一塊放射性金屬不斷衰變,它的重量依公式w=Ae^(-0.05t)在t世紀內由A g減至w g.假設該金屬在1386年內由某重量減至5g.求它原來的重量,準確至最接近的g.

更新1:

一塊放射性物體經t年後的重量w(t) g可由w(t)=Ae^(-ft)得出,其中正值常數f稱為衰變常數.已知鐳要變成原來重量的一半需時1602年.求鐳的衰變常數f,準確至二位有效數字.

更新2:

某城市旅遊的遊客以每年r%的率增長.若遊客在t年內增加了n人,而數目達至N,證明(1+r%)^(-t)=1-n/N.

更新3:

第一題少了每分鐘: 在一個化學反應中,反應物的濃度在5分鐘內由0.037mol/L減至0.012mol/L.假設濃度每分鐘以固定的率減少,8分鐘內濃度會減少至多少??準確至3位小數

回答 (1)

用戶10012

2010-10-26 3:05 am

✔ 最佳答案

1. Concentration dropped from 0.037 to 0.012 in 5 min. and at constant rate, that means, concentration dropped at a constant rate of (0.037 - 0.012)/5 in 1 min. = 0.025/5 = 0.005/min. To drop for a further 3 min., concentration will drop by 0.005 x 3 = 0.015, so concentration becomes 0.012 - 0.015 = - 0.003 (?).
[Could it be drop by 0.012 in 5 min., not drop to 0.012 in 5 min.]
2. w = 5, t = 1386/100 = 13.86
so 5 = A e^(-0.05 x 13.86) = A e^(- 0.693) = 0.5A, so A = 5/0.5 = 10g.
When t = 0, w = A = original weight = 10g.
3. w = half the original value = 0.5A. t in century = 1602/100 = 16.02
so 0.5A = A e^(- f x 16.02)
0.5 = e^( - 16.02f)
ln 0.5 = - 16.02f
- 0.69314 = - 16.02f
f = 0.69314/16.02 = 0.043 ( 2 sig. fig.)
4.
Let original no. of tourist in 1st year = A
so after t years at compound rate, No. of tourist = A(1 + r%)^t = N.....(1)
Since no. of tourist increased = n, that is N - A = n ..............(2)
from (2), A = N - n. Sub. into (1)
(N - n)(1 + r%)^t = N
(N - n)/N = 1/(1 + r%)^t
1 - n/N = (1 + r%)^(-t)

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