Answer :
Answer:
Part a)
W = 10 N
Part b)
h = 1.35 m
Explanation:
Part a)
If object volume is 1 Ltr
density of the object is given as
[tex]\rho = 10^3 \times 1.02[/tex]
[tex]\rho = 1020 kg/m^3[/tex]
now weight of the object is given when it is placed on the table
[tex]W = \rho V g[/tex]
[tex]W = (1020)(10^{-3})(9.81)[/tex]
[tex]W = 10 N[/tex]
Part b)
When pressure due to solution of the mixture is equal to the pressure inside the veins then only it will infuse into the veins
So here we will have
[tex]\rho g h = P_{in}[/tex]
[tex]\rho = 1.03 \times 10^3 kg/m^3[/tex]
[tex]P_o = 13600[/tex]
now we have
[tex]1.03 \times 10^3 (9.81) h = 13600[/tex]
[tex]h = \frac{13600}{1.03 \times 10^3 \times 9.81}[/tex]
[tex]h = 1.35 m[/tex]
This question involves the concepts of specific gravity, density, and guage pressure.
a) The collapsible plastic bag would weigh "10.2 N", if it were resting on a table, out of water.
b) The minimum height of the bag must be "1.35 m" in order to infuse glucose into the vein.
a)
First, we will find the density of the plastic bag using the specific gravity:
[tex]Density\ of\ Liquid = (Specific\ Gravity of\ Liquid)(Density\ of\ Water)\\\rho = (1.02)(1000\ kg/m^3)\\\rho = 1020\ kg/m^3[/tex]
Now, the weight of the bag can be given as follows:
[tex]W=mg\\W=\rho Vg[/tex]
where,
W = Weight = ?
V = Volume of Bag = 1 L = 0.001 m³ (Assumed, missing in question)
Therefore,
[tex]W=(1020\ kg/m^3)(0.001\ m^3)(10\ m/s^2)\\[/tex]
W = 10.2 N
b)
The gauge pressure in the vein is given by the following formula:
[tex]P=\rho gh[/tex]
where,
P = Gauge Pressure = 13600 Pa
[tex]\rho[/tex] = density = (1.03)(1000 kg/m³) = 1030 kg/m³
h = height = ?
Therefore,
[tex]13600\ Pa=(1030\ kg/m^3)(9.8\ m/s^2)h\\\\h=\frac{13600\ Pa}{1030\ kg/m^3(9.8\ m/s^2)}[/tex]
h = 1.35 m
Learn more about density here:
brainly.com/question/24386693?referrer=searchResults
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