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nekit [7.7K]
2 years ago
5

Patient is in the ed due to a football hitting his nose when playing tackle football in the park. X-ray shows a displaced nasal

fracture. What icd-10-cm codes are reported?
Physics
1 answer:
Art [367]2 years ago
6 0

ICD-10-CM codes are -S02.2XXA, W21.01XA, Y93.61, Y92.830


S02.2 for Fracture, Traumatic/Nasal (Bone(s)), ICD-10-CM Alphabetic Index. Both the open fracture code and the dislocation code are not reported. Only the fracture code is provided if a fracture and a dislocation happen at the same place. Search for "dislocation/with fracture" in the alphabetical index to be sent to a doctor. A closed fracture is a fracture with displacement. To report the conditions leading up to the injury, external cause codes are utilized. Look for Struck (accidentally) by/ball (struck) (thrown)/football W21.01 in the ICD-10-CM External Cause of Injuries Index. Seven characters are required in the Tabular List to finish the code. For the first encounter, X is utilized as a stand-in for character number six, and character number seven is given the letter A.

To learn more about ICD-10-CM please visit -brainly.com/question/27932590
#SPJ1

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8 0
2 years ago
Kinematics
leonid [27]

Answer:

a)

a = 2 [m/s^2]

b)

a = 1.6 [m/s^2]

c)

xt = 2100 [m]

Explanation:

In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.

a)

When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:

v_{f} = v_{i}+(a*t)

where:

Vf = final velocity = 40 [m/s]

Vi = initial velocity = 0 (starting from rest)

a = acceleration [m/s^2]

t = time = 20 [s]

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a = 2 [m/s^2]

The distance can be calculates as follows:

v_{f} ^{2} =  v_{i} ^{2}+(2*a*x)

where:

x1 = distance [m]

40^2 = 0 + (2*2*x1)

x1 = 400 [m]

Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.

v = x2/t2

where:

x2 = distance [m]

t2 = 30 [s]

x2 = 40*30

x2 = 1200 [m]

b)

Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.

v_{f} = v_{i}-a*t

Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.

0 = 40 - (a *25)

a = 40/25

a = 1.6 [m/s^2]

The distance can be calculates as follows:

v_{f} ^{2}  = v_{i} ^{2} -2*a*x3\\

0 = (40^2) - (2*1.6*x3)

x3 = 500 [m]

c)

Now we sum all the distances calculated:

xt = x1 + x2 + x3

xt = 400 + 1200 + 500

xt = 2100 [m]

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