Answer:
Explanation:
We know that, V = 140 mL = 0.00014 m3
Assume that it is a sphere. so, we have
V = (4/3) \pir3
r3 = (0.00014 m3) (3) / (4) (3.14)
r = \sqrt[3]{}\sqrt[3]{}3\sqrt{}3.34 x 10-5 m3
r = 1.93 x 10-7 m
(a) The wall tension at end diastole will be given as :
using a formula, we have
T = P r / 2 H
where, P = intraventricular pressure at end diastole = 7 mmHg = 933.2 Pa
H = wall thickness at this time = 0.011 m
then, we get
T = (933.2 Pa) (1.93 x 10-7 m) / 2 (0.011 m)
T = 8.18 x 10-3 N
(b) The wall tension at the end of isovolumetric contraction will be given as :
using a formula, we have
T = P r / 2 H
where, P = intraventricular pressure at end of isovolumetric contraction = 80 mmHg = 10665.7 Pa
H = wall thickness at this time = 0.011 m
then, we get
T = (10665.7 Pa) (1.93 x 10-7 m) / 2 (0.011 m)
T = 9.35 x 10-2 N
(d) The wall stress from A and B which will be given as :
we know that, \sigma = T / w
For part A, we have
\sigmaA = (8.18 x 10-3 N) / (0.011 m)
\sigmaA = 0.743 N/m
For part B, we have
\sigmaB = (9.35 x 10-2 N) / (0.011 m)
\sigmaB = 8.5 N/m
