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3 years ago

Use the drop-down menus to complete the passage.

Physics

2 answers:

user100 [1]3 years ago

8 0

Answer:

The correct answers are as follows:

1. RADIONUCLIDES.

Radionuclides refers to an atom, which has excess energy that makes it to be unstable. The excess energy can be emitted inform of gamma rays, beta particles or alpha particles. These emissions are ionizing radiations which are used in various ways.

2. TRACER.

In nuclear medicine, a tracer refers to a chemical compound that can be used to explore the chemical reaction mechanisms by tracing the path of the chemical reaction. radioactive tracers are usually injected into the blood stream and use to monitor the biochemical reactions taking place in the body.

3. RADIOPHARMACEUTICALS.

Radiopharmaceuticals refers to a group of pharmaceutical drugs which are radioactive in nature. It is usually made up of a radioactive substance and an organic molecule. They are usually given to patients and monitored through imaging equipment for diagnosis or therapeutic purposes.

4. IONIZING RADIATION.

Ionizing radiations refers to those radiations that have the sufficient energy capacity to cause ionization in the medium they pass through. Ionization involves removing an electron from another atom. Ionizing radiations can cause damages to human tissues.

5. DOSAGE.

Ionizing radiation dosage refers to the amount of ionizing radiation that is applied to a particular part of the body. Radiation dose is measured in different ways, which include: absorbed dose, equivalent dose and effective dose. The amount of ionizing dosage given to human should always be considered in order to avoid damage to body tissues.

Explanation:

LenKa [72]3 years ago

4 0

Answer:#1:radionuclides

#2:tracer

#3:radiopharmaceutical

#4:ionizing radiation

#5:dosage

Explanation:

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B. Accuser is called the defendant

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The major evidence for the idea that the expansion of the universe is accelerating comes from observations of:

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The redshirting of the spectra of distant galaxies. They are moving away from us at a rate proportional to their distance

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4 years ago

For pod color, green (G) is dominant over yellow (g). The genotype of a pea plant is Gg. What is the phenotype of this plant?

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Answer:

The correct answer would be green.

It can be explained with the help of law of dominance which states that the dominant trait is expressed in the heterozygous condition.

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3 years ago

Read 2 more answers

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

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a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

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Answers are B,C&D
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