Physics, 29.11.2019 02:31 tmiller2862

a brass statue with a mass of 0.45 kg and a density of 8.00×10^3 kg/m^33 is suspended from a string. when the statue is completely submerged in an unknown liquid, the string tension is 3.4 n. what is the density of the liquid?

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Answer from: mattwe4567893987654

density = 1.96 ₓ 10^3 $\frac{kg}{m^3}$

Explanation:

When a body is immersed in a liquid it exerts an upward force which is equal to the displaced liquid.It is called as "Buoyancy Force".upthrust = dvg where 'd' is density of liquid , 'v' is volume of body, 'g' is gravity.

To find volume of body

v = $\frac{mass}{density}$

v = $\frac{0.45}{8*10^3}$

v = 5.625 ₓ 10^-5 m^3

from FBD,

T = 3.4N = 4.5-upthrust

upthrust = 1.1N = dₓ(5.625 ₓ 10^-5 m^3)ₓ 10m/s^2

d = $\frac{1.1}{5.625}$ ₓ 10^4 $\frac{kg}{m^3}$

d = 1.955555ₓ 10^3 ≅ 1.96 ₓ 10^3 $\frac{kg}{m^3}$

a brass statue with a mass of 0.45 kg and a density of 8.00×10^3 kg/m^33 is suspended from a string.

Answer from: janetexcoelho

density of the statue = 1.832 x $10^{3}$ kg/ $m^{3}$

Explanation:

This question is based on the concept of buoyant force.

Bouyant force is the vertical upward force experienced by a body when it is submerged in a liquid. The magnitude of buoyant force is equal to the weight of the liquid displaced.

Since the body is in equilibrium, the net external force acting on the body is 0.

Here, there are no forces in the horizontal direction. Therefore net external force in the horizontal direction is 0.

Since in the vertical direction also net external force is equal to 0, the total force in the upward direction should be equal to the total force in the downward direction.

Let us first point out the forces acting in this system.

Weight of the statue = mg = 0.45 x 9.8 = 4.41 N (downward)Tension in the string = 3.4 N (upward)Buoyant force = Weight of the liquid displaced (upward)

let the density of the liquid be ρ

Weight of the liquid displaced = (mass of the liquid displaced) x g = (density of the liquid) x (volume of the liquid displaced) x g

volume of the liquid displaced = volume of the statue = $\frac{mass of statue}{density of statue}$ ( since mass = density x volume)

mass of statue = 0.45 kg

density of statue = 8 x $10^{3}$ kg/ $m^{3}$

substituting in the equation gives,

volume of the liquid displaced = $\frac{0.45}{8 X 10^{3} }$ = 5.625 x $10^{-5}$ $m^{3}$

therefore buoyant force = Weight of the liquid displaced = (mass of the liquid displaced) x g = (density of the liquid) x (volume of the liquid displaced) x g = ρ x 5.625 x $10^{-5}$ x 9.8 = ρ x 5.5125 x $10^{-4}$

As we said earlier, since it is in equilibrium,

magnitude of upward force = magnitude of downward force

Tension in the string + Buoyant force = Weight of the statue

substituting,

3.4 + ρ x 5.5125 x $10^{-4}$ = 4.41

ρ x 5.5125 x $10^{-4}$ = 1.01

ρ = $\frac{1.01}{5.5125 X 10^{-4} }$ = 1.832 x $10^{3}$ kg/ $m^{3}$

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