In a major human artery with an internal diameter of 5mm, the flow of blood, averaged over the cardiac cycle is 5cm3·s−1. The ar
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Answer:
see attached
Explanation:
Assuming flow is uniform across the cross section of the artery, the mass flow rate is the product of the volumetric flow rate and the density.
(5 cm³/s)(1.06 g/cm³) = 5.3 g/s
If we assume the blood splits evenly at the bifurcation, then the downstream mass flow rate in each artery is half that:
(5.3 g/s)/2 = 2.65 g/s
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The average velocity will be the ratio of volumetric flow rate to area. Upstream, that is ...
(5 cm³/s)/(π(0.25 cm)²) ≈ 25.5 cm/s
Downstream, we have half the volumetric flow and a smaller area.
(2.5 cm³/s)/(π(0.15 cm)²) ≈ 35.4 cm/s
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Explanation:
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Answer:
It falls at the same speed in both cases.
Explanation:
If I were standing still the phone would be in free fall after slipping out of my hand.
I set a frame of reference with origin on the ground and the positive Y axis pointing up.
It would slip at t0 = 0, from a position Y0 = 5 ft, with a speed of Vy0 = 0.
It would be subject to an gravitational acceleration of -32.2 ft/s^2.
Since acceleration is constant:
Y(t) = Y0 + Vy0 * t + 1/2 * 4 * t^2
When it hits the floor at t1 it will be at Y(t1) = 0
0 = 5 + 0 * t1 - 16.1 * t1^2
16.1 * t1^2 = 5
t1^2 = 5 / 16.1
If the elevator is standing still it would take 0.55 s to hit the ground.
Now, if the elevator is moving up at 10 ft/s.
The frame of reference will have its origin at the place the floor of the elevator is at t = 0, and stay there as the elevetor moves. The floor of trhe elevator will have a position of Ye = 10 * t
Vy0 = 10 ft/s because it will be moving initially at the same speed as the elevator.
And it will hit the floor of the elevator not at 0, but at
Ye = 10 * t2
So:
10 * t2 = 5 + 10 * t2 - 16.1 * t2^2
0 = 5 - 16.1 * t2^2
16.1 * t1^2 = 5
t1^2 = 5 / 16.1
It falls at the same speed in both cases.