Answer:
Explanation:
Formula for Terminal Velocity is given as :
Mass of sugar = Density x Volume
Volume of cube = 5cm x 5cm x 5cm = 125[tex]cm^{3}[/tex] = 0.000125[tex]m^{3}[/tex]
Mass of sugar cube = 1600 kg/[tex]m^{3}[/tex] x 0.000125[tex]m^{3}[/tex] = 0.2 kg
Area = 5cm x 5 cm = 25 [tex]cm^{2}[/tex] = 0.0025 [tex]m^{2}[/tex]
V = [tex]\sqrt{(}[/tex] 2 x 0.2 x 9.81 )/ (1600 x 0.0025 x 0.8)
V = [tex]\sqrt{(}[/tex] 3.924/32) = [tex]\sqrt{0.1226}[/tex]
V= 0.350 m/s
Answer:
The terminal velocity of the sugar would be 40.4 m/s
Explanation:
Given L as the length of the cube = 0.05 m
Area of the sugar = [tex](0.05 m)^{2}[/tex] = 0.0025 [tex]m^{2}[/tex]
The volume of the cube = [tex]L^{3}[/tex] = [tex](0.05 m)^{3}[/tex] = 0.000125 [tex]m^{3}[/tex]
Mass of the sugar = Density x volume =1600 kg/[tex]m^{3}[/tex] x 0.000125
The terminal velocity of a body can be calculated with the expression below;
[tex]V =\sqrt{(2mg)/pAC}[/tex]
Where;
m is the mass of the body = 0.2 kg
g is the acceleration due to gravity = 9.81 m/[tex]s^{2}[/tex]
p is the density of the fluid (air) = 1.2 kg/[tex]m^{3}[/tex]
A is the area of the body = 0.0025 [tex]m^{2}[/tex]
Given L as the length of the cube = 0.05 m
Volume of the cube = [tex]L^{3}[/tex] = [tex]0.05 m^{3}[/tex] = 0.000125 [tex]m^{3}[/tex]
Substituting the values in the equation, we have;
[tex]V =\sqrt{(2*0.2*9.8)/1.2*0.0025*0.8}[/tex]
[tex]V = \sqrt{3.92/0.0024}[/tex]
[tex]V = \sqrt{1633.33}[/tex]
V = 40.4 m/s to the nearest tenth
Therefore the terminal velocity of the sugar would be
40.4 m/s
