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2 years ago

Please solve with explanation

Mathematics

1 answer:

OverLord2011 [107]2 years ago

8 0

Real life scenarios of acute angles are:

Sighting a ball from the top of a building at an angle of 55 degrees.
The angle between two adjacent vanes of a fan that has 6 vanes

<h3>What are acute angles?</h3>

As a general rule, an acute angle, x is represented as: x < 90

This means that acute angles are less than 90 degrees.

<h3>The real life scenarios</h3>

The real life scenarios that involve acute angles are scenarios that whose measure of angle is less than 90 degrees.

Sample of the real life scenarios that satisfy the above definition are:

Sighting a ball from the top of a building at an angle of 55 degrees.
The angle between two adjacent vanes of a fan that has 6 vanes

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3 years ago

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Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

$\frac{dm}{dt} = -\frac{t}{\tau}$

Where:

$m$ - Current mass of the isotope, measured in kilograms.

$t$ - Time, measured in days.

$\tau$ - Time constant, measured in days.

The solution of the differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

$\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}$

$\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }$

The half-life of a isotope ( $t_{1/2}$ ) as a function of time constant is:

$t_{1/2} = \tau \cdot \ln2$

$t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2$

The half-life difference between isotope B and isotope A is:

$\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|$

If $\frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9$ , $t_{A} = 33\,days$ and $t_{B} = 43\,days$ , the difference in the half-lives of the isotopes is: