✔ 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)
