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Zanzabum
2 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]2 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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ale4655 [162]
<span>An object is acted upon by a force of 22 newtons to the right and a force of 13 newtons to the left.

(1)  22 N to the right
(2) 13 N to the left.

magnitude = 22 - 13
magnitude = 9

Direction would be to the right.

So magnitude is 9N direction to the right.</span>
8 0
3 years ago
The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velo
Lina20 [59]
I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J. 
At point A, all this energy has converted into kinetic energy, which is:
K= \frac{1}{2}mv^2
And since K=7.35 J, we can find the velocity, v:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s
3 0
2 years ago
Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13
lyudmila [28]

Answer:

y = 0.14 Cos\left ( 2.512x-34.66t \right )

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

Amplitude, A = 0.14 m

The general equation of the transverse harmonic wave which is travelling right is given by

y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right  )

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

0.14 = 0.14 Sin Ф

Ф = π/2

So, the equation is

y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right  )

y = 0.14 Cos\left ( 2.512x-34.66t \right )

3 0
2 years ago
A rocket fires two engines simultaneously. One produces a thrust of 725Ndirectly forward while the other gives a 513N thrust at
faust18 [17]

The magnitude of the resultant force, F = 1,190.3 acting at a direction X = 13.35°.

<h3>What is the resultant force the two engines exert on the rocket?</h3>

The resultant force on the rocket is calculated thus:

The 513N thrust is resolved into vertical and horizontal components;

Horizontal component: 513N cos(32.4°) = 433.14 N

Vertical component: 513N sin(32.4°) = 274.88 N

Total forward force on the rocket = 725 N + 433.14 N = 1,158.14 N

Total force at right angles:

0 + 274.88 N = 274.88 N

The resultant force (F) is then given as follows:

F² = a² + b²

F² = (1158.14 N)² + (274.88 N)²

F = √1,416,847.27

F = 1,190.3

To find the direction:

tan X 274.88 N / 1,158.14 N

X = tan⁻¹ 0.237346089419241

X = 13.35°

Therefore, the magnitude of the resultant force, F = 1,190.3 acting at a direction X = 13.35°.

In conclusion, the resultant force is obtained by resolving the forces into vertical and horizontal components.

Learn more about resultant force at: brainly.com/question/17434363

#SPJ1

8 0
1 year ago
What do astronomers use in addition to parallax to find the actual distance of stars that are close to Earth?
IgorLugansk [536]

Answer:

trigonometry (guessing)

Explanation:

ellipse: is the shape of an orbit : looks like an oval

periapsis : shortest distance between something like the moon and the planet its orbiting around like the earth

parallax is triangulation. like how gps works. looking at a star one day and then looking at it again 6 months later, an astronomer can see a difference in the viewing angle for the star. With trigonometry, the different angles yield a distance. This technique works for stars within about 400 light years of earth

https://science.howstuffworks.com/question224.htm

By comparing the intrinsic brightness to the star's apparent brightness we can calculate the distance of stars

1/r^2 rule states that the apparent brightness of a light source is proportional to the square of its distance.Jan 11, 2022

https://www.space.com/30417-parallax.html

alternative distance measurement for stars used by most astronomers is the parsec. A star with a parallax angle of 1 arcsecond has a distance of 1 parsec, or 1 parsec per arcsecond of parallax, which is about 3.26 light years

blossoms.mit.edu

.

7 0
2 years ago
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