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Anna [14]
2 years ago
6

After oxygen has been administered, the next priority intervention the nurse would initiate for a patient with a pulmonary embol

us is the administration of which of these therapies?
Physics
1 answer:
dolphi86 [110]2 years ago
3 0

After oxygen has been administered, the next priority intervention the nurse would initiate for a patient with a pulmonary embolus is the administration of IV Heparin.

<h3>What is Heparin?</h3>

Heparin is a prescription medicine used to treat and prevent symptoms of blood clots caused by medical conditions or medical procedures. Heparin may be used alone or with the other medications. Heparin belongs to a class of drugs called the Anticoagulants, Cardiovascular, Anticoagulants, Hematologic.

<h3>What are the side effects of Heparin?</h3>

Heparin may cause serious side effects including:

skin warmth or discoloration,

chest pain,

irregular heartbeats,

shortness of breath,

dizziness,

anxiety,

sweating,

unusual bleeding or bruising,

severe pain or swelling in your stomach, lower back/growing

dark or blue-colored skin on your hands/feet,

nausea,

vomiting,

loss of appetite,

unusual tiredness,

bleeding that will not stop,

nosebleed,

blood in the urine or stool,

black or tarry stools,

coughing up blood or vomit that looks like the coffee grounds,

skin changes where the medication was injected,

fever,

chills,

runny nose,

watery eyes,

easy bruising,

purple or red spots under your skin,

sudden numbness or weakness,

problems with vision or speech, and

swelling or redness in an arm or leg

To know more about cardiovascular visit: brainly.com/question/14349574

#SPJ4

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Answer:

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Let us work with variables and set

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As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).

Now the force of friction is F_s=\mu*N, where N is the normal force and from the diagram it is F_y=mg*cos(\theta).

Thus F_s=\mu*N=\mu*mg*cos(\theta).

Therefore,

F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).

Substituting the value for F_H,m,\mu, and \:\theta we get:

F_{tot}= -38.63N.

Now acceleration is simply

a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.

The negative sign indicates that the acceleration is directed up the incline.

Part B:

d=\frac{1}{2} at^2

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Substitute the value of d=0.50m and a=7.7m/s and we get:

t=0.36s.

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