Answer:
(d) 2 pF
Explanation: the charge on capacitor is given by the expression
Q=CV
where Q=charge
C=capacitance
V=voltage across the plate of the capacitor
here we have given Q=500 pF, V=250 volt
using this formula C=
=500××
=2×
=2 pF
Calculate and plot the radial and circumferential stress distribution in the left ventricle at the end of systole (p 5 80 mmHg;
assume that the ventricle is a spherical shell). The inner radius of the heart is 3.2 cm and the outer radius of the heart is 3.8 cm. The external pressure surrounding the heart is 21 mmHg. Under a disease condition where the heart muscle thickens, calculate the radial and the circumferential stress distribution in the left ventricle at the end of systole. Under these conditions the pressure at the end of systole remains the same, but the inner wall radius is 3 cm and the outer wall radius is 4.2 cm. Compare this to normal conditions and comment.
1 answer:
Answer:
62990.08 N/M
Explanation:
Circumferential stress, or hoop stress, a normal stress in the tangential (azimuth) direction. axial stress, a normal stress parallel to the axis of cylindrical symmetry. radial stress, a stress in directions coplanar with but perpendicular to the symmetry axis.
See attachment for the step by step solution of the given problem..
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Answer:
rate of heat transfer = 9085708.80 W
Explanation:
Given:
Inside diameter, D = 5.1 cm
= 5.1 x m
Average velocity, V = 7 m/s
Mean temperature, T = (66+38) /2
= 52°C
Therefore kinematic viscosity at 52°C is ν = 0.104 X
/ s
Prandtl no., Pr = 0.021
We know Renold No. is
Re =
Re =
= 3.432 X
Therefore the flow is turbulent.
Since the flow is turbulent and the ratio of L/D is greater than 60 we can use Dittua-Boelter equation.
Nu = 0.023 .
= 0.023 x x
= 1221.52
Since Nu =
h =
=
= 225143.3
Therefore rate of heat transfer, q = h.A(T-
q= 225143.3 x 2πrh ( 66-38)
= 225143.3 X 2π X
= 9085708.80 W