Recent content by Gwozdzilla

Homework Statement Blood supply to the heart occurs through coronary arteries. Consider one of the arteries to be 2.5 mm in diameter and 3 cm in length. The average velocity of blood flow through that artery is 1.5 cm/s. Assuming the density of blood to be 1.056 g/cc and viscosity to be 3 cP...

This problem came from a practice exam recommended by my professor that belongs to a course different from, but similar to, what he taught us, so I can't actually speak for the original intentions of this problem. Since I hadn't been taught about friction, I'm fine with keeping it out. I don't...

So the Bernoulli equation is now: 0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2 + fD (L/D)(1/2)ρ(uavg)2 Should the lengths and diameters be averaged too, and then plug in the velocities from my last post to calculate Pinit?

I suppose the tensile force is only being applied in the x direction, whereas forces caused by the mass of the beam would be in the y direction, causing them to not affect each other. But I know I've seen other beam problems before where the beam is fixed to a wall and an object is placed on it...

So if the force is different at each cross sectional area due to the length and mass of the beam, then why isn't the force for the beam with area A1 different from the force for the beam with area A2? Shouldn't this mean that the force providing the tensile stress isn't equal to F?

How do I find pressure without Bernoulli's equation? I've never seen a frictional pressure drop term in Bernoulli's equation; what would that look like?

How would axial stress be calculated? How would this differ from the stress used in the problem?

If flow rate Q = Area * velocity, Q = Au, then A can be redone... A. 0 = ΔP + ρgΔh + (1/2)ρ(Δuavg)2 Pinit = Patm + ρg(hi) + (1/2)ρ(Δuavg)2 where uavg = (1/2)(u1 + u2) u1 = Q1/A1 = Q1/(π(D1/2)2) u2 = Q2/A2 = Q2/(π(D2/2)2) Would this Pinit value be correct? Then for part B, do I just let the...

So when a tensile force is applied to a beam, this force is generally uniform throughout its cross section and does not vary with the length or mass of the beam?

Homework Statement A tank of cross-sectional area A is initially filled with fluid of density ρ and viscosity μ to height hi. The pressure above the fluid in the tank is atmospheric, Patm. At the base of the tank there are two pipes, which are both open to teh atmosphere, such that fluid...

Homework Statement Two beams of different cross sectional areas are fused together to create a single beam, as shown in the figure above. The following relationships apply: A1 = 2A2, L2 = 2L1, E1 = 3E2. A tensile force F is applied at the ends of the bar, as shown. What is the ratio of the...

If I know that ω1 and ω2 are correct by calculating that √ω1ω2 = 1, and I know that I should take the positive values of ω1,2, then why isn't the bandwidth = |ω1-ω2| = |.425-2.355| = |-1.93| = 1.93?

|H(ω)| = ω2/(7ω2 + ω4 + 1) = (1/√2)Hmax = 1/(9√2) Solving for ω gives ω = ±2.355 and ±.425 Bandwidth = B = ω2 - ω1 = 2.78 or 1.93 Is this correct?

|H(ω)| = [(jω)/(3jw + (1-ω2))][(-jω)/(-3jw + (1-ω2))] |H| = ω2/(7ω2 + ω4 + 1) d|H|/dω = 2ω(7ω2 + ω4 + 1)-1 -ω2(7ω2 + ω4 + 1)-2(14ω + 4ω3) = 0 Simplifying gives ω = 1. Is this the center frequency? |H(1)| = 1/9 Is this the bandwidth?

I thought that a bandpass filter was defined as having its |H(ωc)| = 1. Is this not the case?