One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at a

Question

2 years ago

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One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at a

rate of approximately 2.5% per year.
(a) Write the ratio of strontium-90 remaining, p , as a function of years, t , since the nuclear accident.

Mathematics

2 answers:

Art [367] · Answer 1 · 2022-05-15T10:29:07+03:00

The ratio of strontium-90 remaining, p , as a function of years, t , since the nuclear accident is $\boxed{p = {{0.975}^t}}.$

Further explanation:

The exponential decay formula can be expressed as follows,

$\boxed{y = a{{\left( {1 - r} \right)}^x}}$

Here, a represents the initial amount, y is the final amount,r is the rate of decay, and x represents the time.

Given:

The rate of decay is $2.5\%.$

Explanation:

Consider the initial amount of the radioactive substance be $A$ .

The final amount of the radioactive substance is ${{Ap}}$

The ratio of decay can be obtained as follows,

$\begin{aligned}{\text{A}}p &= {\text{A}}{\left( {1 - r} \right)^t}\\p&= \frac{{{\text{A}}{{\left( {1 - r} \right)}^t}}}{{\text{A}}}\\p&= {\left( {1 - 0.025} \right)^t}\\p&= {0.975^t}\\\end{aligned}$

The ratio of strontium-90 remaining, $p$ , as a function of years, $t$ , since the nuclear accident is $\boxed{p = {{0.975}^t}}.$

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Exponential decay

Keywords: twenty percent, contaminants, nuclear accident, Chernobyl, strontium-90, exponentially, 2.5% per year, ratio of strontium, p, function of time, radioactive substance, decays, ten years, each year, rate of decay each year, substance.

igor_vitrenko [27] · Answer 2 · 2022-05-15T08:42:35+03:00

Answer: $P=0.975^t$

Step-by-step explanation:

We know that the general form of the exponential decay formula is

$y=A(1-r)^t$ , where y is final amount remaining after t time, A is the original amount and r is the rate of decay

Now, the ratio of strontium-90 remaining, p , as a function of years, t , since the nuclear accident. $P=\frac{A(1-0.02)^t}{A}=\frac{(0.975)^t}{1}$

Hence, the ratio of remaining since the nuclear accident is $P=0.975^t$