Answer:
3644.63 kg/m^3
Explanation:
mass of rock, m = 1.80 kg
Weight of rock in water = 12.8 N
Weight of rock in water = Weight of rock in air - Buoyant force on the rock due to water
Buoyant force in water = 1.8 x 9.8 - 12.8 = 17.64 - 12.8 = 4.84 N
Volume x density of water x g = 4.84
V x 1000 x 9.8 = 4.84
V = 4.939 x 10^-4 m^3
For the rock to float in a liquid
Buoyant force due to liquid on rock = weight of the rock
V x density of liquid x g = 1.80 x g
density of liquid = 1.80 / (4.939 x 10^-4) = 3644.63 kg/m^3
The smallest density of a liquid in which the rock will float is equal to 3644.63 [tex]kg/m^3[/tex].
Given the following data:
Mass = 1.80 kg.
Weight of string = 12.8 N.
Scientific data:
Acceleration due to gravity (g) = 9.8 [tex]m/s^2[/tex].
Density of water = 1000 [tex]kg/m^3[/tex]
First of all, we would determine the buoyant force that is acting on the rock when it is immersed in water as follows:
Buoyant force = Weight of rock in air - Weight of rock in water
Buoyant force = (1.80 × 9.8) - 12.8
Buoyant force = 17.64 - 12.8
Buoyant force = 4.84 Newton.
Next, we would determine the volume:
Mathematically, the buoyancy of an object is given by this formula;
[tex]F_b = \rho gV\\\\4.84 = 1000 \times 9.8 \times V\\\\V=\frac{4.84}{9800}\\\\V=4.94 \times 10^{-4}\;m^3[/tex]
Now, we can calcuate the smallest density of a liquid:
[tex]F_b = W_a\\\\\rho g V = m g\\\\\rho=\frac{m}{V} \\\\\rho=\frac{1.80}{4.94 \times 10^{-4}}[/tex]
Density = 3644.63 [tex]kg/m^3[/tex].
Read more on density here: brainly.com/question/3173452
