Coronary heart disease and blood pressure
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Coronary heart disease and blood pressure
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[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Coronary heart disease affects 13 millions Americans, and is the leading cause of death both in America and worldwide. Coronary heart disease is caused by a build up of plaque in the arteries that supply blood to the heart. When the flow of blood to the heart is restricted or blocked, the heart can get damaged due to a lack of oxygen. This is commonly referred to as a “heart attack.” Further, if a large piece of plaque buildup gets dislodged from an artery wall, it can get stuck in other arteries throughout the body including the brain.
Consider a person who has a cylindrically shaped piece of plaque that is blocking an artery of radius R as shown in the diagram provided. The blood pressure (gauge pressure) on the right side of the plaque is PR. This particular piece of plaque will break free if the net force on the plaque exerted by blood pressure is greater than Fₒ.
Figure 1. The section of the patient's artery blocked with plaque.
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[question] => How tall would a column of water (density ρw) have to be in order for the bottom of the water column to have the same pressure as the blood on the right side of the plaque?
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[answer] => 1
[description] => Reason for the Correct Answer:
The pressure at the bottom of a column of water is given by Pᴡᴀᴛᴇʀ=ρgh
Pᴡᴀᴛᴇʀ must equal Pʀ
Pʀ=ρᴡgh So h= Pʀ/ρᴡg
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[each_answer] => A. 
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[each_answer] => B. 
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[each_answer] => C. 
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[each_answer] => D. 
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[1] => Array
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[question] => What force does the blood on the right side exert on the piece of plaque?
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[answer] => 1
[description] => Reason for the Correct Answer:
Pressure is defined to be P=F/A
The area of a circle is πr².
P=F/A, so F=PA= PʀπR²
)
[answers] => Array
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[0] => Array
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[each_answer] => A. PʀπR²
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[1] => Array
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[each_answer] => B. Pʀ4R
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[2] => Array
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[each_answer] => C. Pʀ
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[each_answer] => D. Pʀ2πR
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[quiz_unique_key] => 83407773
[question] => Once the plaque clot breaks free, assume it moves with constant velocity to the right along the direction of the artery. What can be said about the pressures if the clot is moving with constant velocity to the right? (Assume no other forces, besides those from the blood pressures, are significant while the clot is moving)
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[answer] => 4
[description] => Reason for the Correct Answer:
ΣF = ma
If the velocity is constant the acceleration is zero
If the acceleration is zero, the net force must be zero
If the net force is zero, the force from the pressures must be equal, so the pressures must be equal
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[each_answer] => A. The force from the pressure on the left side must be smaller than the pressure on the right side, since the clot is moving to the right
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[each_answer] => B. The force from the pressure on the left side must be larger than the pressure on the right side, since a net force is required to keep the clot moving
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[each_answer] => C. The force from the pressure on the left side must be larger than the pressure on the right side, since the clot is moving to the right
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[each_answer] => D. The force from the pressure on the left side must be equal to the pressure on the right side, since the clot is moving with constant velocity
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[question] => If the blood on the right hand side of the plaque exerts a force of magnitude FR, what minimum magnitude of force must the blood on the left side exert on the clot to cause it to dislodge?
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[answer] => 1
[description] => Reason for the Correct Answer:
The net force on the plaque from blood pressure must be at least F𐩒.
Blood is exerting force from the right and left sides of the plaque
Fᴏ = Fʟᴇꜰᴛ- Fʀɪɢʜᴛ so Fʟᴇꜰᴛ = Fᴏ + Fʀ
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[each_answer] => A. Fᴏ + Fʀ
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[each_answer] => B. Fᴏ – Fʀ
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[each_answer] => C. Fᴏ
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[each_answer] => D. 2Fʀ
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[quiz_unique_key] => 574431310
[question] => If the plaque were to slide a distance D to the right, what could be said about the work done on the clot by the lower pressure blood on the right side of the clot?
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[answer] => 2
[description] => Reason for the Correct Answer:
Work = Fdcosθ
The blood on the right side of the block will be exerting a force on the clot to the left
Since the force is in the opposite direction as the direction of motion, the work done is negative since the angle will be 180 degrees and cos180=-1
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[each_answer] => A. The work done by the blood on the right side is positive since the motion of the clot is in the same direction as the force exerted by the blood
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[each_answer] => B. The work done by the blood on the right side is negative since the motion of the clot is in the opposite direction as the force exerted by the blood
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[each_answer] => C. The work done by the blood on the right side is zero since the motion of the clot is in the opposite direction as the force exerted by the blood
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[each_answer] => D. The work done by the blood on the right side is positive since work must always be positive
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