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Zanzabum
3 years ago
5

Certain cancers, especially those of the liver, can be treated by injecting small glass spheres (30 to 35 μm in diameter) contai

ning radioactive 90Y into the tumor. The spheres become lodged in the small capillaries of the tumor, both cutting off its blood supply and delivering a high dose of radiation which damages the tumor, hope-fully sparing surrounding healthy tissue. The spheres stay in place until the radioactivity has completely de-cayed. 90Y undergoes beta-minus decay with a half-life of 64 h; the energy released in the decay is carried away by the beta particle, which has a kinetic energy 0.94 MeV.
a) What is the daughter nucleus for the 90Y decay?
b) The activity is measured when the spheres are prepared; the patient is to be injected 12 days later. If the patient is to be injected with spheres of total activity 1.0 GBq, what should be the approximate activity of the spheres at the time of preparation?
c) Spheres are injected into a liver with a 500 g tumor mass. If the total activity of the spheres administered is 1.0 GBq, and all of the energy deposited ends up in the tumor, what is the dose equivalent in Sv once all of the nuclei have decayed?

Physics
1 answer:
Semenov [28]3 years ago
8 0

Answer:

Explanation:

Base on the scenario been described in the question, let's use the method in the file attached below to solve the question

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Difference between boyles law, charles law, and gay-lussacs law​
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Answer:

All these laws give the relationship between two quantities of the gas among V (volume), p (pressure) and T (temperature), keeping the third one constant - however the two quantities change for each law

Explanation:

Calling:

p = gas pressure

V = gas volume

T = gas temperature (in Kelvin)

We have:

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pV=const.

- Charles's law: the temperature and the volume of a gas kept at constant pressure are directly proportional. Mathematically,

\frac{V}{T}=const.

- Gay Lussac's law: the temperature and the pressure of a gas kept at constant volume are directly proportional. Mathematically,

\frac{p}{T}=const.

3 0
3 years ago
A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling
iogann1982 [59]

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

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6 0
3 years ago
Can you pleas help me!
Sergeeva-Olga [200]

Answer:

T

Explanation:

4 0
4 years ago
Can you answer b-c?
algol [13]
The answer to d is 2.0
5 0
4 years ago
Read 2 more answers
30 unit If the value of an angle with the x-axis of a vector is 70 °, the value of the angle with the y axis is 60, what is the
Nonamiya [84]

Answer: 37.28^{\circ}

Explanation:

Given

Angle made with the x-axis is \alpha=70^{\circ}

Angle made with the y-axis is \beta=60^{\circ}

Suppose angle made with z-axis is \gamma

\therefore \cos^2\alpha+\cos^2\beta+\cos^2\gamma =1

Inserting values

\Rightarrow \cos^2(70^{\circ})+\cos^2(60^{\circ})+\cos^2\gamma=1\\\Rightarrow 0.1169+0.25+\cos^2\gamma=1\\\Rightarrow \cos^2\gamma=0.6331\\\Rightarrow \cos \gamma=0.7956\\\Rightarrow \gamma=37.28^{\circ}

6 0
3 years ago
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