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puteri [66]
3 years ago
13

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.

Engineering
1 answer:
alexandr402 [8]3 years ago
8 0

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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The bulk density of a compacted soil specimen (Gs = 2.70) and its water content are 2060 kg/m^3 and 15.3%, respectively. If the
yaroslaw [1]

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the saturated density should be

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4 0
3 years ago
What is the basic formula for actual mechanical advantage?
slavikrds [6]

Answer:

Mechanical Advantage Formula

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3 0
3 years ago
A Geostationary satellite has an 8kW RF transmission pointed at the earth. How much force does that induce on the spacecraft? (N
soldier1979 [14.2K]

Answer:

The force induced on the aircraft is 2.60 N

Solution:

As per the question:

Power transmitted, P_{t} = 8 kW = 8000 W

Now, the force, F is given by:

P_{t} = Force(F)\times velocity(v) = Fv               (1)

where

v = velocity

Now,

For a geo-stationary satellite, the centripetal force, F_{c} is provided by the gravitational force, F_{G}:

F_{c} = F_{G}

\frac{mv^{2}}{R} = \frac{GM_{e}m{R^{2}}

Thus from the above, velocity comes out to be:

v = \sqrt{\frac{GM_{e}}{R}}

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.979\times 10^{24}}{42166\times 10^{3}}} = 3075.36 m/ s

where

R = R_{e} + H

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where

G = Gravitational constant

T = Time period of rotation of Earth

R is calculated as 42166 km

Now, from eqn (1):

8000 = F\times 3075.36

F = 2.60 N

6 0
3 years ago
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