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I am Lyosha [343]
3 years ago
14

What is the flow of oxygen from inspired air into an alveolar capillary dependent on? A. PO2 in the alveoli being higher than th

e PO2 in the capillary B. PO2 in the alveoli being lower than the PCO2 in the capillary C. PO2 in the alveoli being lower than the PO2 in the capillary D. PO2 in the alveoli being higher than the PCO2 in the capillary
Advanced Placement (AP)
2 answers:
antoniya [11.8K]3 years ago
7 0
I think the correct answer from the choices listed above is option C. The flow of oxygen from inspired air into an alveolar capillary dependent on <span>PO2 in the alveoli being lower than the PO2 in the capillary. Hope this answers the question. Have a nice day.</span>
anygoal [31]3 years ago
4 0

The answer is A. PO2 means the partial pressure of oxygen which is the pressure exerted by an oxygen. Gases diffuse according to their partial pressures. They diffuse from a region of high partial pressure to a region of low partial pressure. In pulmonary exchange, therefore, oxygen with a high partial pressure in the alveoli diffuses into the blood capillaries where PO2 is lower.

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Yes, at a time t such that (√2)/2 ≤ t ≤ 2.

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C(t) = t\times e^{4-t^2} \ for \ 0\leq t\leq 2

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0\times e^{4-0^2} = 0 and 2\times e^{4-2^2} = 2

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\frac{\mathrm{d} \left (t\times e^{4-t^2}   \right )}{\mathrm{d} x}=0

From which we have;

\frac{\mathrm{} e^{4-t^2} - t\times e^{4-t^2} \times2\times t }{(e^{4-t^2} )^2}=0

e^{4-t^2} - t\times e^{4-t^2} \times2\times t }=0

e^{4-t^2}(1 - t\times2\times t })=0\\e^{4-t^2}(1 - 2\times t^2 })=0\\

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For Brandon

For 0 ≤ t ≤ 1, 1 ≤ B(t) ≤8 and for 1 < t ≤ 2, 8 < B(t) ≤ 1.5

1 ≤ f(x) ≤ 1.5

Given that the function B(t) is differentiable, therefore, continuous, there exists a point at which the function C(t) and B(t) intersects given that;

For 0 ≤ t ≤ (√2)/2, 0 ≤ C(t) ≤ 23.416 for (√2)/2 < t ≤ 2, 23.416 > C(t) ≥ 2

and for  0 ≤ t ≤ 0  1 ≤ B(t) ≤ 8 and for 1 < t ≤ 2, 8 > B(t) ≥ 1.5

Therefore, the curves intersect at in between (√2)/2 ≤ t ≤ 2.

8 0
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