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joja [24]
3 years ago
9

A radiographer stands six feet from the x-ray source when performing a portable chest exam and receives an exposure of 2 mGy. If

the radiographer performs a repeat exam using the same technical factors standing at a distance of three feet from the source, how much exposure will be received?
Physics
2 answers:
telo118 [61]3 years ago
3 0

Answer:

  I₂ = 8 mG

Explanation:

The intensity of a beam is

          I = P / A

Where P is the emitted power which is 3) 3

           

Let's use index 1 for the initial position of r₁ = 6 ft and 2 for the second position r₂ = 3 ft

          I₁ A₁  = I₂  A₂

           I₂ = I₁ A₁ / A₂

The area of ​​the beam if we assume that it is distributed either in the form of a sphere is

           A₁ = 4π r²

We substitute

            I₂ = I₁ (r₁ / r₂)²

           I₂ = 2 (6/3)²

           I₂ = 2 4

           I₂ = 8 mG

Mamont248 [21]3 years ago
3 0

Answer:

The amount of exposure that will be received at 3 ft is 8 mGy

Explanation:

Here, we note that the  amount of radiation exposure of the radiographer is given by the inverse square law. That is the amount of radiation exposure is directly proportional to the inverse square of the distance that is

\frac{Old \, \, Intensity}{New \, \, \, Intensity} = \frac{(New\, distance)^2}{(Old\, distance)^2} \therefore \frac{2}{New \, \, \, Intensity} = \frac{3^2}{6^2}

Or New intensity = 2\times \frac{36}{9}  = 8mGy

Therefore, the amount of exposure that will be received at 3 ft = 8 mGy.

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A .2kg Basketball is pitched with a velocity or 40 m/s and then bed and into the picture with a velocity of 60 m/s. What is the
Tamiku [17]

Answer:

40kgm

Explanation:

∆p = m(v - u)

= 2(60 - 40)

= 2 × 20

= 40kgm/s

4 0
3 years ago
A machinist with normal vision has a near point at 25 cm. The machinist wears eyeglasses in order to do close work. The power of
Alik [6]

Answer:

17.4 cm

Explanation:

Power of lens = +1.75 diopters

Focal length of lens

f=\frac{100}{1.75}=\frac{400}{7}

This is a convex lens as focal the diopter given is positive which makes the focal length positive. Image distance will be negative.

v = -25

\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\Rightarrow \frac{1}{-25}-\frac{1}{u}=\frac{1}{\frac{400}{7}}\\\Rightarrow -\frac{1}{u}=\frac{7}{400}+\frac{1}{25}\\\Rightarrow \frac{1}{u}=-\frac{7}{400}-\frac{1}{25}\\\Rightarrow \frac{1}{u}=-0.0575\\\Rightarrow u=-17.4\ cm

∴ The new near point is 17.4 cm

7 0
3 years ago
Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

3 0
3 years ago
Which of the following would produce the most power?
Fantom [35]

Answer:

A mass of 10 kilograms lifted 10 meters in 5 seconds.

Explanation:

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

Power = \frac {Energy}{time}

But Energy = mgh

Substituting into the equation, we have

Power = \frac {mgh}{time}

Given the following data;

Mass = 10kg

Height = 10m

Time = 5 seconds

We know that acceleration due to gravity is equal to 9.8 m/s²

Power = \frac {10*9.8*10}{5} = 490 Watts

Hence, a mass of 10 kilograms lifted 10 meters in 5 seconds would produce the most power.

6 0
2 years ago
Find the velocity and distance travelled of a car that is accelerating at 3 m/s2 for 4 seconds.
marta [7]

Explanation:

Given:

v₀ = 0 m/s

a = 3 m/s²

t = 4 s

Find: Δx and v

Δx = v₀ t + ½ at²

Δx = (0 m/s) (4 s) + ½ (3 m/s²) (4 s)²

Δx = 24 m

v = at + v₀

v = (3 m/s²) (4 s) + 0 m/s

v = 12 m/s

6 0
2 years ago
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