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3 years ago

Which radioisotope is used in dating geological formations?

Physics

1 answer:

ziro4ka [17]3 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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What is the speed of a wave on a string with a wavelength of 1.75 m and a frequency of 2.0 Hz

deff fn [24]

Answer:

V=3.5 m/s

Explanation:

V=(F)(W)

V=(2)(1.75)

V= 3.5 m/s

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3 years ago

Introduced species often thrive and multiply in an environment very different from their original one. Why are they often able t

Anestetic [448]

They begin to adapt into their new location. They then end up having adaptations to help them survive.

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3 years ago

Suppose a woman raises a 65 N object in 2m in 4 seconds.

Novosadov [1.4K]

Answer:

$\huge\boxed{\sf P.E = 130\ Joules}$

$\huge\boxed{\sf P = 32.5\ Watts}$

Explanation:

<u>Given Data:</u>

Weight = W = 65 N

Height = h = 2 m

Time = t = 4 secs

<u>Required:</u>

Power = P = ?

Work Done in the form of Potential Energy = P.E. = ?

<u>Formula:</u>

P.E. = Wh

P = P.E. / t

<u>Solution:</u>

P.E. = (65)(2)

P.E = 130 Joules

P = P.E. / t

P = 130 / 4

P = 32.5 Watts

$\rule[225]{225}{2}$

Hope this helped!

8 0

3 years ago

Your cell phone typically consumes about 390 mW of power when you text a friend. If the phone is operated using a lithium-ion ba

yulyashka [42]

Answer:

I = 0.11 A

Explanation:

In an electric circuit, the power delivered to a load, is just the product of the potential difference between the load terminals, times the current flowing through it, as follows:

$P = V*I$

In this case, the power is the one consumed by the cell phone = 390 mW, and the voltage the one produced by the internal energy of the battery, 3.5 V, neglecting the voltage loss at the internal resistance of the battery.
So, we can solve the above equation for the current I, as follows:

$I = \frac{P}{V} = \frac{0.39W}{3.5V} = 0.11 A$

The current flowing through the cell-phone circuitry is 0.11 A.

4 0

3 years ago

Is this right?? please help me. IT IS SOCIOLOGY!!

Westkost [7]

Answer:

Yes, it is correct : )

Explanation:

Hope this helps!

8 0

2 years ago

Read 2 more answers