What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface
1 answer:
Answer:
The density of the woman is 950.8 kg/m³
Explanation:
Given;
fraction of the woman's volume above the surface = 4.92%
then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%
the specific gravity of the woman
The density of the woman is calculate as;
Density of fresh water = 1000 kg/m³
Density of the woman = 0.9508 x 1000 kg/m³
Density of the woman = 950.8 kg/m³
Therefore, the density of the woman is 950.8 kg/m³
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Missing details: figure of the problem is attached.
We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,
where:
is the pressure difference between the two ends
is viscosity of the fluid
L is the length of the pipe
is the volumetric flow rate, with
being the section of the tube and
the velocity of the fluid
r is the radius of the pipe.
We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:
is the dynamic water viscosity at
L=4.0 cm=0.04 m
and r=1 mm=0.001 m
Using these data in the formula, we get:
However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.
