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You are correct that mass density $\rho$ is given by mass $m$ per volume $V$:

\(\displaystyle \rho=\frac{m}{V}\tag{1}\)

which thus implies:

\(\displaystyle m=\rho V\tag{2}\)

So, to determine the density of the object, find its mass using the given density of water and the volume of water displaced in (2), then use this value for the mass to determine the object's density using the amount of coconut oil it displaced as its volume in (1).

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In water

You are correct that mass density $\rho$ is given by mass $m$ per volume $V$:

\(\displaystyle \rho=\frac{m}{V}\tag{1}\)

which thus implies:

\(\displaystyle m=\rho V\tag{2}\)

So, to determine the density of the object, find its mass using the given density of water and the volume of water displaced in (2), then use this value for the mass to determine the object's density using the amount of coconut oil it displaced as its volume in (1).

\(\displaystyle m =1000 kg m^{-3} * 23 cm^3 \tag{2}\)

\(\displaystyle m =23000 kg\)

Using coconut oil

\(\displaystyle m =900 kg m^{-3} * 25 cm^3 \tag{2}\)

\(\displaystyle m = 22500 kg \tag{2}\)

Where should the acceleration due to gravity be used mentioned in the problem $g=10 ms^{-2}$

Something still looks wrong here

Many Thanks

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- Mar 5, 2012

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Hey mathlearn !In water

\(\displaystyle m =1000 kg m^{-3} * 23 cm^3 \tag{2}\)

\(\displaystyle m =23000 kg\)

Using coconut oil

\(\displaystyle m =900 kg m^{-3} * 25 cm^3 \tag{2}\)

\(\displaystyle m = 22500 kg \tag{2}\)

Where should the acceleration due to gravity be used mentioned in the problem $g=10 ms^{-2}$

Something still looks wrong here

Many Thanks

I think your units are a little off - those masses are on par with trucks -

We don't need the acceleration due to gravity.

It's probably provided because Archimedes' Principle says it's the upward

So formally we should include $g$, but it will be divided out again, so we don't need to know its value.

Oh, and that irregularly shaped object could almost be the votive crown of Hiero of Syracuse, which triggered Archimedes to make his discovery, yell Eureka (twice), and run naked into the streets of Syracuse.

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- Mar 5, 2012

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The volume of the votive crown (jk) is:Hmmm a demonstration would be helpful here

$$V=25 \text{ cm}^3$$

due to the results of its immersion in coconut oil.

Its mass is:

$$m = 1000 \text{ kg/m}^3 \times 23 \text{ cm}^3

= 1000 \text{ kg/m}^3 \times 23 \cdot 10^{-6} \text{ m}^3

= 23 \cdot 10^{-3} \text{ kg} = 23\text{ g}

$$

due to its immersion in water.

So the density is:

$$\rho = \frac mV = \frac{23\text{ g}}{25 \text{ cm}^3} = \frac{23}{25} \frac{\text{g}}{\text{cm}^3} = \frac{23}{25} \frac{\text{kg}}{\text{L}}$$

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Yes. And if it floats, we can still measure the volume - by pushing it under.Many Thanks So As Mark said the density of the object was less than of water and was greater than of coconut oil slightly

So to determine the volume of that votive crown it should be fully submerged in some liquid like for instance in coconut oil here ?