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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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Dvinal [7]
I attached the full question.
We know that for a parallel-plate capacitor the surface charge density is given by the following formula:
\sigma=\varepsilon_0 \frac{V}{d}
Where V is the voltage between the plates and d is separation.
Voltage is by definition:
V=Ed
Voltage is analog to the mechanical work done by the force.
Above formula is correct only If the field is constant, and we can assume that it is since no function has been given.
The charge density would then be:
\sigma=\varepsilon_0 \frac{Ed}{d}=\varepsilon_0E\\
\sigma= 8.85\cdot10^{-12}\cdot 2.1\cdot 10^6= 0.0000185\frac{c}{m^2}
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6 0
4 years ago
An object with velocity 141 ft/s has a kinetic energy of 1558.71 ft∙lbf, on a planet whose gravity is 31.5 ft/s2. What is its
Sidana [21]

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m  = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \  \times \ \ \ \ 1   \ lbf\ }

m  = \frac{(2 \ \times \ 1558.71  \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71  \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm

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6 0
3 years ago
What happens when the voltage increases and the resistance stays the same in a electrical circuit?
Orlov [11]

Answer:

The current in the circuit increases

Explanation:

The ohm's law states that the potential across a circuit is proportional to the current in the circuit.

                                             V ∝ I

Where 'V' is the potential difference across the circuit and 'I' is the current in the circuit.

The proportionality constant present in the equation is the resistance of the circuit. Hence, the equation becomes

                                             V = IR

According to the equation, when V is directly proportional to 'I' where 'R' remains as constant, then the change in 'V is brings change in 'I' to make the equation valid.

So, when there is an increase in the voltage, the current on the circuit increases.

4 0
3 years ago
What must the diver's minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, wh
Dafna1 [17]

Answer:

1.08 m/s

Explanation:

This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.

Time taken to fall 9.5 m

vertical acceleration = a = 9.8 m/s^2.

vertical velocity = 0, (since there is only horizontal component for velocity, )

distance traveled  s = 9.5 m.

Substituting these values in the equation

s= u \timest+0.5at^{2}

t= \sqrt{\frac{2s}{g} }

t=\sqrt{\frac{2\times9.5}{9.8} }

⇒ t= 1.392 sec

Velocity needed

We know the time taken (1.392 s) to travel 1.5 m,

So velocity = 1.5 m / 1.392 s = 1.08 m/s

hence velocity of the diver must be at least 1.08 m/s

4 0
3 years ago
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mixas84 [53]
The answer would be C. 5m
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Hope this helps!
8 0
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