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3 years ago

Which statement is true about nuclear decay

Biology

1 answer:

Readme [11.4K]3 years ago

5 0

Answer:

Nuclear decay rates are constant

Explanation:

Spontaneous decay of unstable nuclei is a process of a statistical nature, it is not possible to predict with certainty when an individual radioactive nucleus will decay, nor can the decay process be influenced in any way.

Therefore, the half-life of radionuclides (specific for each radionuclide) is defined as the time period in which half of the initial number of radioactive nuclei decays.

The unit of measurement for the radioactivity of the material is the becquerel (Bq) and denotes the number of decays in one second (1Bq = 1 decay / 1 second).

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Letter B (found in the image above) represents a cell in which phase?

RSB [31]

Anaphase because the chromosomes are moving away (Ana- Away)

6 0

3 years ago

Read 2 more answers

Which of the three traits considered in this film (bipedality, extensive tool use, and large brains) were present in the 3.2-mil

belka [17]

Answer:

The bipedality

Explanation:

<u>One of the things the discovered fossil signified was that human bipedality was more ancient than the large brain size</u> because Lucy actually had a small skull which could indirectly be translated to small brain size.

NOTE: Bipedality can be described to mean the ability of an organism to move about with two legs. Hence, it must have been discovered that Lucy had two legs.

8 0

3 years ago

What are some examples of index fossils

nata0808 [166]

Answer:

Ammonites, , Brachiopod, Grapolithina, Trilobites and Nanofossils are examples of index fossils

Explanation:

Ammonites were common during the Mesozoic Era (245 to 65 mya), but were not found after the Cretaceous period. They went extinct during the K-T extinction (65 mya)

Brachiopods which are mollusk-like marine animals appeared during the Cambrian (540 to 500 mya). Some types of them still survive

Grapolithina which are widespread colonial marine hemichordates that also lived during the Cambrian period.

Nanofossils are microscopic fossils which are remains of calcareous nannoplankton, coccolithophores from various eras

Trilobites were common during the Paleozoic Era (540 to 245 mya); about half of the Paleozoic fossils are Trilobites. Trilobites evolved at the beginning of the Paleozoic Era and went extinct somewhere in the late Permian period which was 248 million years ago

This might be extra information but i hope it helps

6 0

3 years ago

How are sendimentary rocks formed

Sonja [21]

Answer:

Sedimentary rocks are formed when sediment is deposited out of air, ice, wind, gravity, or water flows carrying the particles in suspension. This sediment is often formed when weathering and erosion break down a rock into loose material in a source area.

Explanation

8 0

3 years ago

A research team studies a large Eastern Milk Snake population in a metropark. In these snakes, gene T is one gene that affects t

sweet [91]

Answer:

Option E

Explanation:

As per Hardy Weinberg's second equation of equilibrium, the sum of genotype frequency for all possible genotype in a given population with two allele is equal to

$p^2 + q^ 2+2pq = 1\\$

Substituting the given values in above equation, we get -

$0.16 + 0.64 + 0.20 =1\\ p^2+q^2+2pq= 1$

Hence, second equation satisfies.

As per Hardy Weinberg's first equation of equilibrium, sum of all allele frequencies is equal to 1

$p + q=1\\\sqrt{0.16} +\sqrt{0.64} \\0.4+0.8 = 1.2\neq 1$

Hence, this population is not in Hardy weinberg's equation.

After one generation, the frequency of T allele is $0.4$ which is the same as that of frequency of T allele in previous generation

Also as per hardy Weinberg's theory , if a population genotype frequency sums up to 1, it is in equilibrium if its allele frequency sum is also 1.

But here in 2004. This was not the case.

However, Weinberg has stated that the population which is not in equilibrium can reach equilibrium after one generation only with a new set of genotype frequency and allele frequencies.

Hence, in this case there is a possibility that allele and genotype frequencies have readjusted to sum up to 1

If frequency of the T allele is $0.4$

New frequency of the t allele would be

$1- 0.4\\= 0.6$

and New genotype frequencies would be

$p^ 2 = 0.4^2 = 0.16\\q^ 2 = (1-0.4)^2 = 0.36\\2pq = 1-0.16-0.36 = 0.48$

Thus, $p^ 2+q^2+2pq = 1$

Hence, option E is correct

5 0

3 years ago