The half-life of cesium-137 is 30 years. Suppose that we start with 60 grams of cesium-137 in a storage pool. How many half-lives will it take for there to be 10 grams of cesium-137 in the storage pool? (Round answer to two decimal places.)

How many years is that? (Round answer to one decimal place.)

9514 1404 393

Explanation:

The remaining amount, in terms of t half-lives, is ...

  q(t) = 60·(1/2)^t

We want t when q(t) = 10, so ...

  10 = 60·(1/2)^t

  1/6 = (1/2)^t . . . . . . divide by 60

  log(1/6) = t·log(1/2) . . . . take logs

  t = log(1/6)/log(1/2) = -0.778151/-0.30103 ≈ 2.58496

It will take about 2.58 half-lives for there to be 10 grams remaining.

In years, that is 2.58×30 = 77.5 years.

the body's situation won't change the forces are equal opposit and on the same body

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.