# A 3.40 kg aluminum ball has an apparent mass of 2.10 kg when submaged in a particular liquid. What is the density of the liquid?

As you know, when an object is *submerged* in a particular liquid, the difference between its mass and its apparent mass will represent the mass of the liquid displaced by the object.

You can think of mass in terms of density and *volume*. More specifically, you know that the following relationship exists between mass, volume, and density

#color(blue)("density" = "mass"/"volume")#

This means that the mass of the displaced liquid will be equal to

#rho_"liquid" = m_"liquid"/V_"liquid" implies m_"liquid" = rho_"liquid" * V_"liquid"#

But the volume of the displaced liquid will be equal to the volume of the ball,which in turn can be written as

#V_"ball" = m_"ball"/rho_"ball"#

Plug this into the above equation to get

#m_"liquid" = rho_"liquid" * m_"ball"/rho_"ball"#

We've already established that the mass of the displaced liquid is the difference between the mass and the apparent mass of the ball, which means that you have

#overbrace(m_"ball" - m_"apparent")^(color(red)(m_"liquid")) = m_"ball" * rho_"liquid"/rho_"ball"#

This means that the density of the liquid will be

#rho_"liquid" = rho_"ball" * (m_"ball" - m_"apparent")/m_"ball"#

Now all that you need is the density of aluminium, which is listed as

#rho_(Al) = "2712 kg/m"^3#

https://tutor.hix.ai

Plug in your values and find the density of the liquid

#rho_"liquid" = "2712 kg/m"^3 * ( (3.40 - 2.10) color(red)(cancel(color(black)("kg"))))/(3.40 color(red)(cancel(color(black)("kg"))))#

#rho_"liquid" = color(green)("1040 kg/m"^3)#

The answer is rounded to three sig figs.

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The liquid has a density of 9.5 kg/L.

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The apparent mass of an object submerged in a liquid is equal to the difference between the mass of the object and the mass of the liquid displaced by the object, which is equal to the volume of the object multiplied by the density of the liquid and the acceleration due to gravity.

Given:

- Mass of the aluminum ball, ( m_{\text{ball}} = 3.40 ) kg
- Apparent mass of the aluminum ball in the liquid, ( m_{\text{apparent}} = 2.10 ) kg
- Density of aluminum, ( \rho_{\text{aluminum}} = 2700 ) kg/m(^3)
- Acceleration due to gravity, ( g = 9.81 ) m/s(^2)

The volume of the aluminum ball can be calculated using its mass and density:

[ V_{\text{ball}} = \frac{m_{\text{ball}}}{\rho_{\text{aluminum}}} ]

The apparent mass of the aluminum ball in the liquid is equal to the mass of the aluminum ball minus the mass of the displaced liquid:

[ m_{\text{apparent}} = m_{\text{ball}} - \rho_{\text{liquid}} \cdot V_{\text{ball}} \cdot g ]

Substituting the expressions for ( V_{\text{ball}} ) and ( m_{\text{apparent}} ) and solving for ( \rho_{\text{liquid}} ) gives:

[ \rho_{\text{liquid}} = \frac{m_{\text{ball}} - m_{\text{apparent}}}{V_{\text{ball}} \cdot g} ]

[ \rho_{\text{liquid}} = \frac{m_{\text{ball}} - m_{\text{apparent}}}{\frac{m_{\text{ball}}}{\rho_{\text{aluminum}}} \cdot g} ]

[ \rho_{\text{liquid}} = \frac{m_{\text{ball}} - m_{\text{apparent}}}{m_{\text{ball}} / \rho_{\text{aluminum}} \cdot g} ]

Calculating ( \rho_{\text{liquid}} ) using the given values:

[ \rho_{\text{liquid}} = \frac{3.40 , \text{kg} - 2.10 , \text{kg}}{3.40 , \text{kg} / 2700 , \text{kg/m}^3 \cdot 9.81 , \text{m/s}^2} ]

[ \rho_{\text{liquid}} = \frac{1.30 , \text{kg}}{0.927 , \text{m}^3} ]

[ \rho_{\text{liquid}} = 1.402 , \text{kg/m}^3 ]

Therefore, the density of the liquid is ( 1.402 , \text{kg/m}^3 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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