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timurjin [86]
3 years ago
8

((Bruv pls hurry on test)) Radioisotopes often emit alpha particles, beta particles, or gamma rays. The distance they travel thr

ough matter increases in order from alpha to gamma. Each radioisotope has a characteristic half-life, which is the time needed for half of a sample of radioisotope to undergo nuclear decay.
Which quality is desirable for a radioisotope that is used for medical imaging of a specific organ? Check all that apply.

half-life of several years
half-life of several days
half-life of several seconds
emission of gamma rays
emission of alpha particles
ability to be attached to a compound used by the body
ability to pass through the body without being absorbed
Physics
2 answers:
Jlenok [28]3 years ago
3 0

Answer:

A,B,,G

Explanation:

got it right on edg

mihalych1998 [28]3 years ago
3 0

Answer:

2,4,6

Hey I actually saw this answer in the comment section of the incorrect answer to this question and put it up here so people can verify it. Well, hope this helps dude:)

You might be interested in
Please Someone Help me I Need Help ASAP PLEASE
dimaraw [331]
The match making one is Radiation is i think B and Convection i think is C and Conduction i think is A.  I think 1 is radiation. Because radiation is transferred between objects or an empty space. Number 2 i think is conduction Because conduction is the transfer of thermal energy between to objects touching. Number 3 i also i think is radiation because radiation transfers between objects. Number 4 i also think is conduction because the spoon was touching a hot pot that was on the stove so then the thermal energy was transferred between the two objects touching. number 5 I think is convection because convection is the transfer of energy by the movement of a fluid, such as air or water. Number 6 i think is radiation because the snake is not touching the lamp and radiation goes all across spaces. Number 7 i think is convection because convection is the movement of a fluid, such as air or water. Hope this helps sorry its so long :)      
4 0
3 years ago
How is a controlled variable different from a responding variable?
Arisa [49]

Answer:

The answer is a controlled variable stays the same throughout an experiment, but a responding variable changes

Explanation:

just did it on apex

5 0
2 years ago
50 points !! I need help asap.......Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to t
r-ruslan [8.4K]

1) At the top of the building, the ball has more potential energy

2) When the ball is halfway through the fall, the potential energy and the kinetic energy are equal

3) Before hitting the ground, the ball has more kinetic energy

4) The potential energy at the top of the building is 784 J

5) The potential energy halfway through the fall is 392 J

6) The kinetic energy halfway through the fall is 392 J

7) The kinetic energy just before hitting the ground is 784 J

Explanation:

1)

The potential energy of an object is given by

PE=mgh

where

m is the mass

g is the acceleration of gravity

h is the height relative to the ground

While the kinetic energy is given by

KE=\frac{1}{2}mv^2

where v is the speed of the object

When the ball is sitting on the top of the building, we have

  • h=40 m, therefore the potential energy is not zero
  • v=0, since the ball is at rest, therefore the kinetic energy is zero

This means that the ball has more potential energy than kinetic energy.

2)

When the ball is halfway through the fall, the height is

h=20 m

So, half of its initial height. This also means that the potential energy is now half of the potential energy at the top (because potential energy is directly proportional to the height).

The total mechanical energy of the ball, which is conserved, is the sum of potential and kinetic energy:

E=PE+KE=const.

At the top of the building,

E=PE_{top}

While halfway through the fall,

PE_{half}=\frac{PE_{top}}{2}=\frac{E}{2}

And the mechanical energy is

E=PE_{half} + KE_{half} = \frac{PE_{top}}{2}+KE_{half}=\frac{E}{2}+KE_{half}

which means

KE_{half}=\frac{E}{2}

So, when the ball is halfway through the fall, the potential energy and the kinetic energy are equal, and they are both half of the total energy.

3)

Just before the ball hits the ground, the situation is the following:

  • The height of the ball relative to the ground is now zero: h=0. This means that the potential energy of the ball is zero: PE=0
  • The kinetic  energy, instead, is not zero: in fact, the ball has gained speed during the fall, so v\neq 0, and therefore the kinetic energy is not zero

Therefore, just before the ball hits the ground, it has more kinetic energy than potential energy.

4)

The potential energy of the ball as it sits on top of the building is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

h = 40 m is the height of the building, where the ball is located

Substituting the values, we find the potential energy of the ball at the top of the building:

PE=(2)(9.8)(40)=784 J

5)

The potential energy of the ball as it is halfway through the fall is given by

PE=mgh

where:

m = 2 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

h = 20 m is the height of the ball relative to the ground

Substituting the values, we find the potential energy of the ball halfway through the fall:

PE=(2)(9.8)(20)=392 J

6)

The kinetic energy of the ball halfway through the fall is given by

KE=\frac{1}{2}mv^2

where

m = 2 kg is the mass of the ball

v = 19.8 m/s is the speed of the ball when it is halfway through the  fall

Substituting the values into the equation, we find the kinetic energy of the ball when it is halfway through the fall:

KE=\frac{1}{2}(2)(19.8)^2=392 J

We notice that halfway through the fall, half of the initial potential energy has converted into kinetic energy.

7)

The kinetic energy of the ball just before hitting the ground is given by

KE=\frac{1}{2}mv^2

where:

m = 2 kg is the mass of the ball

v = 28 m/s is the speed of the ball just before hitting the ground

Substituting the values into the equation, we find the kinetic energy of the ball just before hitting the ground:

KE=\frac{1}{2}(2)(28)^2=784 J

We notice that when the ball is about to hit the ground, all the potential energy has converted into kinetic energy.

Learn more about kinetic and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0
3 years ago
Is calculating the change of velocity the same as calculating acceleration? ​
koban [17]

Answer:

Yes! Thinking about it graphically a position vs time graph models meters per second in most cases, making every point on the line have the units m/s. If we want the find the slope we are finding the change between each point and those units would change to m/s/s or m/s^2 giving us the same units for acceleration. Simply put, slope of a velocity graph gives us acceleration.

Explanation:

4 0
2 years ago
Consider the following True/False statements:
Ainat [17]

Answer:

6) False

7) True

8) False

9) False

10) False

11) True

12) True

13) True

14) True

Explanation:

The spacing between two energy levels in an atom shows the energy difference between them. Clearly, B has a greater value of ∆E compared to A. This implies that the wavelength emitted by B is greater than A while B will emit fewer, more energetic photons.

When atoms jump from lower to higher energy levels, photons are absorbed. The kinetic energy of the incident photon determines the frequency, wavelength and colour of light emitted by the atom.

The energy level to which an atom is excited is determined by the kinetic energy of the incident electron. As the voltage increases, the kinetic energy of the electron increases, the further the atom is from the source of free electrons, the greater the required kinetic energy of free electron. When electrons are excited to higher energy levels, they must return to ground state.

4 0
3 years ago
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