One more study problem to help for physics test?

Question

In an aortic aneurysm, a bulge forms where the walls of the aorta are weakened. If blood flowing through the aorta (radius 1.0 cm) enters an aneurysm with a radius of 3.1 cm, how much on average is the blood pressure higher inside the aneurysm than the pressure in the unenlarged part of the aorta? The average flow rate through the aorta is 110 cm3/s. Assume the blood is non-viscous and the patient is lying down so there is no change in height.
___ Pa

Answer 1

I. calculate the velocity of blood flow in the aorta and in the aneurysm.


II. apply Bernouli's equation (and disregarding elevation):

V1^2 / 2 +P1/rho = Constant=V2^2 / 2 +P2/rho

(V1 and P1 are the velocity and pressure in the aorta, and V2 and V2 are the velocity and pressure in the aneurysm.)


III. therefore

P2-P1 = rho(V1^2 - V2^2)

Pressure rise = rho(V1^2 - V2^2)

Plug n Chug

Answer 2

they're telling you that there is no vertical element of preliminary velocity and additionally you're attempting to come across the horizontal ingredient. by way of fact the diver's horizontal velocity won't replace as quickly as she leaves the cliff, you may desire to make certain how long she has in the previous she could hit the ledge. then you definately understand how briskly she needs to be going while she leaves the cliff with a view to not hit it. you may ignore approximately her weight - the acceleration of gravity is a persevering with. i could initiate with the help of enforcing a coordinate gadget on the concern. i could placed the muse on the lip of the cliff. Now the ledge is at -9.00m. you may desire to understand while she'll be on the point of the ledge. First, make certain the equation expression her vertical place. y = -one million/2*g*t^2. Set that equation equivalent to -9 and remedy for t. i visit apply 9.8 m/s^2 for g (it extremely is the acceleration of gravity, and a few training will use 10 m/s^2.) i'm getting one million.355 seconds. Now you recognize how long she has to return and forth one million.75m with a view to dodge the ledge. basically divide one million.75m with the help of one million.355 seconds for the horizontal element of velocity. i'm getting one million.29 m/s.