# Answer to Question #6702 in Mechanics | Relativity for yah

Question #6702

the pressure inside a 1.0cm radius vessel carrying blood with a speed of 0.50m/s is 5200 Newton/ square meter, while the pressure outside is 3200 Newton/ square meter. to what radius must the vessel be reduced so that the outside pressure is greater than that inside, thus making the vessel close at the constriction?

Expert's answer

Let's find the area of the vessel:

S = 4πr² = 4π1² = 4π.

We denote the inside pressure force by Fi and outside pressure force by Fo. When the radius is r=1 the values of these forces are:

Fi = S*5200 = 4π5200 = 20800π,

Fo = S*3200 = 4π3200 = 12800π.

We see that Fi > Fo.

We know that inside pressure force is constant under the volume changes. Let's denote the new radius of the vessel by R.

Fi = Fo

20800π = 3200*4πR² ==> R = sqrt(20800/4/3200) = sqrt(1.625) = 1.2747

So, the vessel must expand a bit, then the outside pressure will grow up.

S = 4πr² = 4π1² = 4π.

We denote the inside pressure force by Fi and outside pressure force by Fo. When the radius is r=1 the values of these forces are:

Fi = S*5200 = 4π5200 = 20800π,

Fo = S*3200 = 4π3200 = 12800π.

We see that Fi > Fo.

We know that inside pressure force is constant under the volume changes. Let's denote the new radius of the vessel by R.

Fi = Fo

20800π = 3200*4πR² ==> R = sqrt(20800/4/3200) = sqrt(1.625) = 1.2747

So, the vessel must expand a bit, then the outside pressure will grow up.

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