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Irina18 [472]
3 years ago
10

Describe the path of blood through the heart, starting at the superior and inferior vena cava and leaving the aorta.

Physics
1 answer:
Svet_ta [14]3 years ago
4 0

The superior and inferior vena cava are the two main blood vessels that bring de-oxygentated blood back to the heart after the blood's done a circuit of the body to offload its oxygen to organs and muscles and tissues and cells, etc. The blood from these vessels enters the right atrium of the heart. Once the atrium is full of blood it contracts, pushing the blood through the open tricuspid valve into the right ventricle of the heart. The tricuspid valve closes shut once the blood is in the right ventricle, preventing the blood from returning to the right atrium. While the valve is closed, the ventricle contracts, pushing the blood out of the heart through the pulmonic valve, into the pulmonary artery, where it is carried directly to the lungs where it gets a new supply of oxygen attached to it. The freshly oxygenated blood courses along the pulmonary vein from the lungs back to the heart, entering the heart's left atrium. When the left atrium is full of blood, it empties by contraction, through the open mitral valve, into the left ventricle of the heart. The mitral valve then closes, so the blood cannot retrace its path. The left ventricle then contracts, pushing the blood out of the heart through the aortic valve into the big blood vessel called the aorta, from where it courses round the body delivering oxygen again.

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Read 2 more answers
Juan is on a morning jog. His speed is represented in the graph. At what rate of speed is Juan running between 4min and 6min? Ac
Nesterboy [21]

Answer:

<em>a) The speed of Juan is 250 m/min </em>

<em>b) In that interval, he's at rest </em>

<em>c) Juan would have needed 12 minutes to run 3000 m</em>

Explanation:

<u>Distance vs Time Graph Analysis </u>

If time is on the horizontal axis, and the distance is in the vertical axis, then the slope of the graph represents the instantaneous speed of the moving object. The graph shows three different zones which reflect Juan's jogging at that morning.

a) The period between 4 and 6 minutes clearly shows Juan is moving in such a way the distance increases when time passes. The speed of the motion can be obtained by computing the slope of the line. We can locate the points (4,1000) and (6,1500). We compute the speed  

\displaystyle v=\frac{1500-1000}{6-4}=\frac{500}{2}=250 m/min

The speed of Juan is 250 m/min

b) We can see between 7 and 11 minutes, Juan's distance is not changing because he stopped running. In that interval, he's at rest

c) We have already determined the speed on the first 7 minutes (the same as between 4 and 6 minutes). We know that

\displaystyle v=\frac{x}{t}

Where v is the speed, x is the distance, and t is the time

We need to know the time needed to travel x=3000 m at the initial speed v=250 m/min, so we solve the equation for t

\displaystyle t=\frac{x}{v}

\displaystyle t=\frac{3000}{250}=12\ minutes

Juan would have needed 12 minutes to run 3000 m

3 0
4 years ago
Explain why a moving object cannot come to a stop instantaneously (in zero seconds)
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Explanation:

Knowledge

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3 years ago
What is the change in momentum of a 4 kg object accelerating from 10 m/s to 12 m/s? In this problem what is the change in veloci
Ksju [112]

Answer:

8 kg*m/s and 2m/s

Explanation:

momentum is mass times velocity. since the velocity changes by 2 m/s. you just take the difference from final velocity to beginning velocity.

4kg*12m/s = 48 kg*m/s

4kg*10m/s = 40 kg*m/s

48kg*m/s - 40 kg*m/s = 8kg*m/s

7 0
3 years ago
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