Answer:
See the answers below.
Explanation:
We must apply the definition of density which is the ratio of mass to volume.
[tex]Ro=m/v[/tex]
where:
Ro = density = 2560 [kg/m³]
m = mass [kg]
v = volume [m³]
With the column height data and column radius we can calculate the volume, since it resembles the volume of a cylinder. In this way the volume of a cylinder is defined by the following expression.
[tex]V=\pi *r^{2}*h[/tex]
where:
r = radius = 45 [cm] = 0.45 [m]
h = height = 2.72 [m]
[tex]V = \pi *(0.45)^{2}*2.72 \\V = 1.73 [m^{3} ][/tex]
a)
The mass can be calculated as follows.
[tex]m=Ro*V\\m=2560*1.73\\m= 4428.8 [kg][/tex]
ii)
Pressure is defined as the relationship of force over an area.
The area can be calculated with the radius of the cylinder.
[tex]A=\pi *r^{2}\\A = \pi *(0.45)^{2}\\A = 0.636 [m^{2} ][/tex]
Therefore
[tex]P=F/A\\where:\\F=m*g\\F=4428.8*9.81\\F=43446.53[N][/tex]
[tex]P=43446.53/0.636\\P=68312.15[Pa]\\P=68.31[kPa][/tex]
iii)
If the pressure is very large we must analyze the pressure equation again.
[tex]P=\frac{F}{A}[/tex]
We can see that the pressure depends on the strength (weight) and the area of the column. In this way if we increase the value of the area (denominator) we see that the pressure decreases.
If the new Radius is 50 [cm] = 0.5 [m]
[tex]A=\pi *(0.5)^{2}\\A = 0.785 [m^{2} ][/tex]
Now we can calculate again the pressure.
[tex]P=43446.53/0.785\\P=55317.84[Pa]=55.32[kPa][/tex]
We can see the reduction in pressure.
