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Airida [17]
3 years ago
7

Which applications, either for diagnostic purposes or for therapeutic purposes, do not involve ionizing radiation? Check all tha

t apply. CT scan MRI ultrasound laser surgery radiopharmaceuticals PET scan
Physics
2 answers:
Umnica [9.8K]3 years ago
8 0

Answer:

Applications in which ionizing radiation are not used are  

 MRI

ultrasound

 laser surgery

Explanation:

Ionizing radiation

Ionizing radiations are radiation which have sufficien energy to ionize an atom. Example of ionizing radiation are x rays, gamma rays and all particle radiation from radioactive decay

CT Scan

CT scan is an abbreviated form of computerized tomography. CT scan provide more information about a body than x-rays. In this test a sequence of x-ray images are taken at various angle of a body and these images after computer processing  provide information about the body. Since x-ray is ionizing radiation so in CT scan ionizing radiations are used.

Ultrasound

In ultra sound scan high frequency sound waves(which are not ionizing radiation) are used to find the information about a body.

MRI

MRI is an abbreviated form of Magnetic resonance imaging. In MRI scanning high intensity magnetic field and radio waves(which are not ionizing radiation) are used to find the information of a body.

Laser surgery

It is a type of surgery in which laser beam(which are not ionizing radiation) is used to cut the skin or to remove a lesion from skin surface.

Radiopharmaceuticals PET scan

PET scan is an abbreviated form of positron emission tomography scan. This test is used to find the images of a body to detect certain diseases like cancer and heart diseases in your body.  In this test radioactive tracers are used. In this test ionizing radiation are used.

neonofarm [45]3 years ago
5 0

These applications DO NOT INVOLVE harmful ionizing energy:

- MRI

- ultrasound

- laser surgery

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What is the magnification when an object is placed at 2f from the pole of the convex mirror? 
Gekata [30.6K]

Answer:

Linear magnification = 1/3

Explanation:

Given:

Convex mirror

Object's distance from pole = 2f

Find:

Linear magnification

Computation:

Object distance, u = −2f

So,

1/v + 1/u = 1/f

1/v + 1/(-2f) = 1/f

1/v = 1/f + 1/2f

BY taking LCM

1/v = 3 / 2f

v = 2f / 3

Magnification, M = -v / u

So,

Magnification, M = (2f / 3) / 2f

Magnification, M = 2f / 6f

Magnification, M = 2 / 6

Linear magnification = 1/3

 

5 0
3 years ago
A 1000 Kg car traveling at 10 m/s hits the back of a 5000 Kg truck
insens350 [35]

Answer:

0 m/s

The car becomes stationary

Explanation:

The law of conservation of linear momentum states that the sum of inital and final momentum should be equal and expressed as

m_cu_c+m_tu_t=m_cv_c+m_tv_t

Where m represent the mass, u and v are tge initial and final velocities while subscripts c and t represent car and truck.

Taking forward direction as positive then considering that the truck is originally at rest, we substitute original truck velocity with 0, mass of car and truck with 1000 kg and 5000 kg respectively then final truck velocity as 2 m/s as we take initial car velocity to be 10 m/s

1000*10+(5000*0)=5000*2+1000v

1000v=0

V=0

Therefore, the car finally becomes stationary.

8 0
3 years ago
Block A of mass 2.0 kg is released from rest at the top of a 3.6 m long plane inclined at an angle of 30o, as shown in the figur
Jet001 [13]

Answer:

............................

31kg

4 0
3 years ago
Read 2 more answers
Calculate the index of refraction for a medium in which the speed of light is 2.1x 108 m/s. The speed of light in vacuum is 3x10
strojnjashka [21]

Answer:

n = 1.42

Explanation:

The refractive index for a medium is given by the ratio of the speed of light in vacuum to the speed of light in a medium.

n=\dfrac{c}{v}\\\\n=\dfrac{3\times 10^8}{2.1\times 10^8}\\\\n = 1.42

So, the refractive index of the medium is 1.42.

5 0
3 years ago
A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr
nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

 a= μ g                          ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g  =ω²X

We know for spring mass system natural frequency given as

\omega=\sqrt{\dfrac{k}{M+m}}

By putting the values

\omega=\sqrt{\dfrac{150}{5.67+1}}

ω = 4.47 rad/s

μ g  =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2

kX^2=(m+M)v^2

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0
3 years ago
Read 2 more answers
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