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Irina-Kira
1 month ago
12

Bradley gets an x-ray at a radiology clinic that employs its own technologists and radiologists. Would the coder at the clinic r

eport the technical and professional components of this x-ray separately? Would the answer differ if Bradley’s x-ray were sent to an independent radiologist for interpretation? Would any modifiers be used?
Physics
1 answer:
Maru [3.3K]1 month ago
8 0

Answer:

Explanation:

If Bradley's examination was done and analyzed in the same facility, the radiologist code is utilized as shown for example- procedure code 72100- Radiologic examination, spine, lumbosacral, 2 or 3 views is reported.

if the X-ray was conducted by Dr. X but he doesn't interpret the results and instead passes it on to the radiologist for initial assessment, then a 26-modifier is applied. For instance, a report from the technologist would be procedure code 72050-Radiologic examination, spine, cervical, 2 or 3

views or under specific circumstances, 72050-TC and the consulting radiologist could report 72050-26.

if Bradley’s x-ray were referred to an independent radiologist for interpretation, then procedure code 76140 would be used in the reporting.

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The fundamental equation is derived from Mr. Planck: E=h \nu, where h is Planck’s constant and ν is the frequency. This relationship describes the energy per photon at a specific frequency. Although a wavelength is provided, it can easily be converted to frequency using the equation: c= lambda / nu, where c denotes the speed of light; λ (lambda) is the wavelength; and ν is the frequency. Once the energy of a photon with a wavelength of 550nm is determined, it will show how many photons are needed to gather 10^-18J. Remember to pay attention to the units.
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1 month ago
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A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it
Ostrovityanka [3204]

Response:D

Clarification:

Provided

mass of object m=5 kg

Distance traveled h=10 m

resulting velocity v=12 m/s

energy conservation occurs starting when the object begins its descent and reaches a speed of 12 m/s

Initial Energy=mgh=5\times 9.8\times 10=490 J

Final Energy=\frac{1}{2}mv^2+W_{f}

=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}

where W_{f} is the work done by friction, if any

490=360+W_{f}

W_{f}=130 J

As friction is present, this indicates an open system with a net external force of zero.

An open system allows for the exchange of energy and mass, and the presence of friction indicates that it is indeed an open system.

4 0
2 months ago
Steve and Elsie are camping in the desert, but have decided to part ways. Steve heads north, at 8 AM, and walks steadily at 2 mi
Keith_Richards [3271]

Answer:

2.57 hours

Explanation:

Let t (in hours) represent the time it takes for Elsie to walk until they are separated by 25 miles. Since Steve starts 2 hours earlier, his time will be t + 2.

The distance covered by Steve heading north is s_s = 2(t + 2)

The distance Elsie travels heading west is s_e = 2.5t

The separation between Steve and Elsie is

\sqrt{s_s^2 + s_e^2} = \sqrt{(2(t+2))^2 + (2.5t)^2} = 25

We can solve for t by squaring both sides.

(2(t+2))^2 + (2.5t)^2 = 25^2 = 625

4(t+2)^2 + 6.25t^2 = 625

4(t^2 + 4t + 4) + 6.25t^2 = 625

10.25t^2 + 16t - 609 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{-16\pm \sqrt{(16)^2 - 4*(10.25)*(-109)}}{2*(10.25)}

t= \frac{-16\pm68.74}{20.5}

Therefore, t = 2.57 or t = -4.13.

Since t cannot be negative, we select t = 2.57 hours.

8 0
2 months ago
A book is pushed with an initial horizontal velocity of 5.0 meters per second off the top of a 1.19 meter high desk. How far awa
Yuliya22 [3333]

Answer:

2.45 m

Explanation:

To begin, we need to determine the book's time of flight, utilizing the equation for vertical motion:

h=\frac{1}{2}gt^2

with

h = 1.19 m representing the height traveled by the book

g = 9.8 m/s^2 being the gravitational acceleration

t symbolizing the time of flight

By solving for t,

t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.19)}{9.8}}=0.49 s

Next, we calculate the horizontal distance the book travels, defined by

d=v_x t

where

v_x = 5.0 m/s denotes the horizontal velocity

t = 0.49 s being the time of flight

By substituting,

d=(5.0)(0.49)=2.45 m

Thus, the book lands at a distance of 2.45 m.

8 0
2 months ago
Part A
kicyunya [3294]

Answer:

v' = -18 m/s

Explanation:

  • Assuming no external forces act during the collision, the overall momentum must be preserved as follows:

       p_{o} = p_{f} (1)

  • The initial momentum can be defined this way (with the initial direction of the ball considered positive):

       m_{b} * v_{b} -M_{c}*V_{c} = m_{b} * 18 m/s + (-M_{c}* 20 m/s) (2)

  • The final momentum can be formulated as follows (given that v'b moves in the opposite direction to vb):

        -(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)

  • If we consider that Mc >> mb, we assume the car's speed remains unchanged as a result of the collision, allowing us to substitute V'c for Vc in (3).
  • Consequently, we can rewrite (3) like this:

       -(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (4)

  • By substituting (2) and (4) back into (1), we arrive at:

       m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (5)

  • By simplifying and rearranging, we can solve for v'b like this:
  • v'_{b} = -18 m/s (6), which makes sense because it's as though the ball collided with a wall and simply reversed its velocity post-collision.
3 0
1 month ago
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