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2 years ago

Which radioisotope is used in dating geological formations?

Physics

1 answer:

ziro4ka [17]2 years ago

6 0

<h2>Answer: U-238</h2>

Explanation:

Let's begin by explaining that for radioactive geological dating (also called radioisotope dating) in which radioactive impurities were selectively incorporated when the fossil materials were formed, it is very useful to compare it with a naturally occurring radioisotope having a known half-life.

Now, taking into account that the <u>fossils are millions and millions of years old, radioisotopes are needed that exceed this measure. </u>

To understand it better:

The longer the half-life of a radioisotope, the greater its utility for estimating fossil ages or geological formations.

In this sense, uranium-238 (U238) has a half-life of 4,470 million years, therefore, it is among the most commonly used radioisotopes for fossil and geological dating.

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Car A hits car B (initially at rest and of equal mass) frombehind while going 35 m/s. Immediately after the collision, car Bmove

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Answer:

The fraction of kinetic energy lost in the collision in term of the initial energy is 0.49.

Explanation:

As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.

By definition, the kinetic energy is:

k = 0.5*mV^2

Expressing the initial and final kinetic energy for cars A and B:

$ki=0.5*maVa_{i}^2+0.5*mbVb_{i}^2$

$kf=0.5*maVa_{f}^2+0.5*mbVb_{f}^2$

Since the masses are equals:

$m=ma=mb$

For the known velocities, the kinetics energies result:

$ki=0.5*mVa_{i}^2$

$ki=0.5*m(35 m/s)^2=612.5m^2/s^2*m$

$kf=0.5*mbVb_{f}^2$

$kf=0.5*m(25 m/s)^2=312.5m^2/s^2*m$

The lost energy in the collision is the difference between the initial and final kinectic energies:

$kl=ki-kf$

$kl = 612.5m^2/s^2*m-312.5 m^2/s^2*m=300 m^2/s^2*m$

Finally the relation between the lost and the initial kinetic energy:

$kl/ki = 300 m^2/s^2 * m / 612.5 m^2/s^2 * m$

$kl/ki = 24/49=0.49$

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