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MaRussiya [10]
3 years ago
13

Which applications, either for diagnostic purposes or for therapeutic purposes, involve radioactive materials placed in the body

? Check all that apply.
radiopharmaceuticals
MRI
laser eye surgery
external beam radiation therapy
lithotripsy
brachytherapy
radioisotope imaging
Physics
2 answers:
CaHeK987 [17]3 years ago
5 0

When a radioactive material is required to be placed in the body, the applications are brachytherapy and radioisotope imaging.

Radioactive materials are elements which has the ability to disintegrate by emitting radioactive substance or radiation. A good example of this is Cobalt-60, Titanium-99 etc.

Brachytherapy is a therapeutic process in which radioactive material is inserted into the body in close proximity to the region affected. The radioactive material emits radiations which are required to control the unwanted biological material in the body. A good application of this is the treatment of cancer using Cobalt-60.

Radioisotope imaging is a diagnostic process which is an imaging technique that may require placing a radioactive material in the body so as to trace or locate the affected part of the body. In this case, the material is used as a tracing element.

The applications that require the placing of radioactive materials in the body are brachytherapy and radioisotope imaging.

For more explanation, visit: brainly.com/question/9790340

vitfil [10]3 years ago
3 0

Answer:

radiopharmaceuticals

brachytherapy

radioisotope imaging

Explanation:

edge 2020

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a ball of diameter 10 cm and mass 10 grams is dropped in a container of water. the cross sectional area of the container is 100
Anastaziya [24]

Answer:

h = 9.83 cm

Explanation:

Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water

       

let's reduce the magnitudes to the SI system

         r = 10 cm = 0.10 m

         m = 10 g = 0.010 kg

         A = 100 cm² = 0.01 m²

the definition of density is

          ρ = m / V

the volume of a sphere

         V = \frac{4}{3} \ \pi r^{3}

          V = \frac{4}{3} π 0.1³

          V = 4.189 10⁻³ m³

let's calculate the density of the ball

           ρ = \frac{0.010}{4.189 \ 10^{-3} }

           ρ = 2.387 kg / m³

the tabulated density of water is

         ρ_water = 997 kg / m³

we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation

            B - W = 0

            B = W

             

where B is the thrust that is given by Archimedes' principle

           ρ_liquid  g V_submerged = m g

           V_submerged = m / ρ_liquid

we calculate

            V _submerged = 0.10 9.8 / 997

             V_submerged = 9.83 10⁻⁴ m³

The volume increassed of the water container

           V = A h

            h = V / A

let's calculate

            h = 9.83 10⁻⁴ / 0.01

            h = 0.0983  m

this is equal to h = 9.83 cm

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