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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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Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti
Alex

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = m_1=6.50 kg- 1.55 kg=4.95 kg

Velocity achieved by squid = v_1=2.60 m/s

Momentum gained by squid = P=m_1v_1

Mass of water = m_2=1.55 kg

Velocity by which water was released by squid = v_2

Momentum gained by water but in opposite direction = P'=m_2v_2

P = P'

m_1v_1=m_2v_2

v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E  =\frac{1}{2}m_1v_1^{2}

Kinetic energy of water = K.E' = \frac{1}{2}m_2v_2^{2}

Total kinetic energy created by this maneuver:

K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}

=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules

4 0
3 years ago
a passenger elevator operates at an average of 8 m/s if the 60th floor is 219 m above the first floor how long does it take the
MAVERICK [17]

Answer:

The universal sign for choking is __________.

A.

two balled fists pressing the abdomen

B.

pointing at an open mouth

C.

two hands grasping the neck

D.

pretending to cough

Explanation:

4 0
3 years ago
A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.
bija089 [108]

a)

i) Potential for r < a: V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

ii) Potential for a < r < b:  V(r)=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

iii) Potential for r > b: V(r)=0

b) Potential difference between the two cylinders: V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c) Electric field between the two cylinders: E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

E=\frac{\lambda}{2\pi \epsilon_0 r}

where

\lambda is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

E=0

So the potential where the electric field is zero is constant:

V=const.

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density +\lambda and an equal negative charge density -\lambda. Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)

However, we know that the potential at b is zero, so

V(r)=V(b)=0

ii) The electric field in the region a < r < b instead it is given only by the positive charge +\lambda distributed over the surface of the inner cylinder of radius a, therefore it is

E=\frac{\lambda}{2\pi r \epsilon_0}

And so the potential in this region is given by:

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r} (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

And so, for r<a,

V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

b)

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

- Potential at the surface of the outer cylinder:

V(b)=0

Therefore, the potential difference is simply equal to

V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c)

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

The electric field is just the derivative of the electric potential:

E=-\frac{dV}{dr}

so we can find it by integrating the expression for the electric potential. We find:

E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
Are the objects described here in static equilibrium, dynamic equilibrium, or not equilibrium at all? Explain.
Alexandra [31]
Let us examine the given situations one at a time.

Case a. A 200-pound barbell is held over your head.
The barbell is in static equilibrium because it is not moving.
Answer: STATIC EQUILIBRIUM

Case b. A girder is being lifted at a constant speed by a crane.
The girder is moving, but not accelerating. It is in dynamic equilibrium.
Answer: DYNAMIC EQUILIBRIUM

Case c: A jet plane has reached its cruising speed at an altitude.
The plane is moving at cruising speed, but not accelerating. It is in dynamic equilibrium.
Answer: DYNAMIC EQUILIBRIUM

Case d: A box in the back of a truck doesn't slide as the truck stops.
The box does not slide because the frictional force between the box and the floor of the truck balances out the inertial force. The box is in static equilibrium.
Answer: STATIC EQUILIBRIUM
4 0
3 years ago
Read 2 more answers
Select the correct answer.
kvasek [131]

Answer:

8.37×10⁻⁴ N/C

Explanation:

Electric Field: This is the ratio of electrostatic force to electric charge. The S.I unit of electric field is N/C.

From the question, the expression for electric field is given as,

E = F/Q.......................... Equation 1

Where E = Electric Field, F = force experienced by the charged balloon, Q = Charge on the balloon.

Given: F = 8.2×10⁻² Newton, Q = 9.8×10 Coulombs = 98 Coulombs

Substitute these values into equation 1

E = 8.2×10⁻² /98

E = 8.37×10⁻⁴ N/C

Hence the Electric Field of the charged balloon =  8.37×10⁻⁴ N/C

4 0
3 years ago
Read 2 more answers
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