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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:)

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Question 3 (5 points)
Art [367]

Answer:

There no image

Explanation:

7 0
3 years ago
Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration
eimsori [14]

The complete text of the problem is:

<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>

<em />

<u>Solution:</u>

1) Acceleration: -2336 m/s^2 on the hardwood floor, -382 m/s^2 on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

v^2 - u^2 = 2ad

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

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

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

d = 1.8 mm = 0.0018 m

So we can find the acceleration by using again the equation

v'^2 - v^2 = 2ad

Solving for a,

a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2

For the carpeted floor instead,

d=1.1 cm = 0.011 m

therefore the acceleration is

a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

a=\frac{v'-v}{t}

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

a=-2336 m/s^2

So the duration of the collision is

t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms

For the carpeted floor,

a=-382 m/s^2

So the duration of the collision is

t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

8 0
3 years ago
The near point of an eye is 110 cm. A corrective lens is to be used to allow this eye to focus clearly on objects 26.0 cm in fro
irga5000 [103]

Answer:

The focal length of the lens is 34.047 cm

The power of the needed corrective lens is 2.937 diopter.

Explanation:

Distance of the object from the lens,u = 26 cm

Distance of the image from the lens ,v= -110 cm

(Image is forming on the other side of the lens)

Since ,lens of the human eye is converging lens,convex lens.

Using a lens formula:

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{26 cm}+\frac{1}{-110 cm}

f = 34.047 cm = 0.3404 m

Power of the lens = P

P=\frac{1}{f}=\frac{1}{0.34047 m}=2.937 diopter

6 0
2 years ago
PLEASE HELP ASAP!!!
musickatia [10]

Answer:

22.2 W

Explanation:

First of all, we calculate the work done by moving the wagon, using the formula:

W=Fdcos \theta

where

F = 20 N is the magnitude of the force

d = 1000 m is the displacement of the wagon

\theta=0^{\circ} is the angle between the direction of the force and of the displacement (assuming the force is applied in the direction of motion)

Substituting, we find

W=(20)(1000)=20,000 J

Now we can find the power generated, which is equal to the ratio between the work done and the time taken:

P=\frac{W}{t}

where

W = 20,000 J

t = 15 min = 900 s

Substituting,

P=\frac{20000}{900}=22.2 W

And the same value in Joules/second (remember that 1 Watt = 1 Joule/second)

5 0
3 years ago
What value do we use to describe acceleration due to gravity
andrey2020 [161]

Answer:

9.8 m/s2

Explanation:

In the first equation above, g is referred to as the acceleration of gravity. Its value is 9.8 m/s2 on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is 9.8 m/s2.

Got it from the internet, hope it helps though ^^

7 0
3 years ago
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