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Orlov [11]
2 years ago
8

Identify how scientists use radioactive isotopes by selecting the best answers from the drop-down menus. Archaeologists use radi

oactive isotopes to ___________ Scientists use radioactive isotopes in agriculture to_____________ Geologists use radioactive isotopes to ______________
Physics
2 answers:
salantis [7]2 years ago
6 0

Answer:

Archaeologist use radioactive isotopes to determine the ages of various objects, rocks and materials. This is called radioactive dating. Radioactive isotope Carbon-14 is widely used for this dating process.

Scientists use radioactive isotopes in agriculture to monitor or study the uptake and use of essential nutrients by plants from the soil. This helps to determine viability, productivity and nutritious ability of the plants on a piece of land.

Geologists use radioactive isotopes to trace leaks in underground water storage, pipes. Radioactive isotopes are effective tracers because their radioactivity can be easily detected.

SpyIntel [72]2 years ago
5 0

Answer:

Explanation:

Archaeologists use radioactive isotopes for Radiometric dating which is typically a method of dating based on the rate of decay of radioactive isotopes present in all organic materials.

The radiometric dating technique tyat is Carbon-14 or radiocarbon is used widely in archaeology.

Scientists use radioactive isotopes in agriculture as research tools in the development of new strains of agricultural crops that are drought and disease resistant, of higher quality, have shorter growing time and produce a higher yield of produce.

Radioactive elements can be used to understand numerical age of geological materials like rock ages, composition, origin etc. on time scales as long as (and even longer than) the age of the Earth. The use of Radiometric dating is employed.

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ddd [48]

Explanation:

Given that,

Initial speed of the electron, u=4\times 10^5\ m/s

Distance, s = 5 cm = 0.05 cm

Acceleration of the electron, a=6\times 10^{12}\ m/s^2  

(a) Let v is the electron's velocity when it emerges from this region. It can be calculated as :

v^2=u^2+2as

v^2=(4\times 10^5)^2+2\times 6\times 10^{12}\times 0.05

v = 871779.788 m/s

or

v=8.71\times 10^5\ m/s

(b) Let t is the time for which the electron take to cross the region. It can be calculated as:

t=\dfrac{v-u}{a}

t=\dfrac{8.71\times 10^5-4\times 10^5}{6\times 10^{12}}

t=7.85\times 10^{-8}\ s

Hence, this is the required solution.

4 0
2 years ago
You are working on a laboratory device that includes a small sphere with a large electric charge Q. Because of this charged sphe
madam [21]

Answer:

the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.

Explanation:

We can answer this exercise using Gauss's law

      Ф = ∫ e . dA = q_{int} / ε₀

field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell.  the flow must be zero since the charge of the sphere is equal  induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field

From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.

5 0
2 years ago
In the lab setup above, block 1 is being pulled across a smooth surface by a light string connected across a pulley to a falling
Nuetrik [128]

Answer:

As block 1 moves from point A to point B, the work done by gravity on block 2 is equal to the change in the kinetic energy of the two-block system.

Explanation:

As block 2 goes down , work is done by gravity on block 2 . This is converted

into kinetic energy of block 1 and block 2 . Work done by gravity is mgh which can be measured easily . kinetic energy of both the blocks can also be measured.  

8 0
2 years ago
A point charge q1 = 4.50 nC is located on the x-axis at x = 1.95 m , and a second point charge q2 = -6.80 nC is on the y-axis at
Vinvika [58]

Answer:

Explanation:

One charge is situated at x = 1.95 m . Second charge is situated at y = 1.00 m

These two charges are situated outside sphere as it has radius of .365 m with center at origin. So charge inside sphere = zero.

Applying Gauss's theorem

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= 0 Ans .

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2 years ago
Differences between Pressure and upthrust​
Angelina_Jolie [31]

Answer:

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4 0
2 years ago
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