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MaRussiya [10]
2 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]2 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]2 years ago
3 0

Answer:

radiopharmaceuticals

brachytherapy

radioisotope imaging

Explanation:

edge 2020

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Explanation:

A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

           F = m a  

           - e E = m a

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with the field directed downward, the acceleration is in the vertical upward direction.

We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

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           x = v₀ t

           t = x / v₀

y axis (perpendicular to plates)

          y = y₀ + v_{oy} t + ½ a t²

Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known

          y = ½ a t²

we substitute

         y = ½ (e E /m)  (x / v₀)²

         y = ½ e x2 /m v₀²   E

we look for the electric field

        E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

let's calculate

         E = 2  9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

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B) The electron is exchanged for a proton

Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

          y = ½ e x² / m v₀² E

          y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)²   145.6

          y = 28.4375 10⁻⁸ m

since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small

In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)

C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

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           F_{g} = G m M / R²

electron

          Between the electron and the positive charges of the conducting plate

           F_{g}= 6.67 10⁻¹¹ 1.67 10⁻²⁷ 9.1 10⁻³¹ / (0.5 10⁻³)²

           F_{g} = 4.1 10⁻⁵¹ N

           

electric force

           F_{e} = -e E

           F_{e} = - 1.6 10⁻¹⁹ 145.6

           F_{e} = 2.3 10⁻¹⁷ N

let's look for the reason between these two forces

         F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹

          F_{e} /F_{g} = 10³⁴

We see that the electric force is many orders of magnitude higher than the gravitational force.

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