You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 N. If you then lower the statue into a tub of water so that it is completely submerged, the scale reads 17.0 N. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution.

When the tub is lowered and submerged in water, the scale reads 17N. So the statue is subjected to a force that make the difference of 28.4 - 17 = 11.4N. This equals to the gravity force of water displaced.

[tex]\rho_wVg = 11.4[/tex]

Let water density [tex]\rho_w = 1000kg/m^3[/tex], we can calculate the volume of the water displaced, which is also the volume of the statue:

[tex]V = \frac{11.4}{g\rho_w} = \frac{11.4}{9.8*1000} = 0.00116 m^3[/tex]

The density of the statue is mass divided by its volume:

[tex]\rho = \frac{m}{V} = \frac{2.9}{0.00116} = 2491 kg/m^3[/tex]

hamzaahmeds hamzaahmeds · Answer 2 · 2021-11-04T12:28:20+01:00

This question involves the concepts of density, weight, volume, and buoyant force.

The density of the ceramic statue is "2495.5 kg/m³".

First, we will find out the mass of the statue:

[tex]W = mg\\m=\frac{w}{g}[/tex]

where,

W = hanging weight of statue = 28.4 N

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]m =\frac{28.4\ N}{9.81\ m/s^2}\\[/tex]

m = 2.9 kg

Now, we will find out the volume of the statue. The difference, in weight of the statue upon submerging, must be equal to the buoyant force applied by the water. This buoyant force is equal to the weight of the volume of water displaced, which is equal to the volume of the statue.

[tex]Difference\ in\ weight\ of\ statue=(Density\ of\ water)(Volume\ of\ Statue)g\\28.4\ N-17\ N=(1000\ kg/m^3)(V)(9.81\ m/s^2)\\\\V=\frac{11.4\ N}{(1000\ kg/m^3)(9.81\ m/s^2)}[/tex]

V = 1.16 x 10⁻³ m³

Now, the density of the ceramic is given as follows:

[tex]\rho = \frac{m}{V} = \frac{2.9\ kg}{1.16\ x\ 10^{-3}\ m^3}\\\\\rho=2495.5\ kg/m^3[/tex]

Learn more about buoyant force here:

https://brainly.com/question/21990136?referrer=searchResults

The attached picture illustrates the buoyant force.