The density (more precisely, the volumetric mass density; also known as specific mass), of a substance is its mass per unit volume. The symbol most often used for density is ρ (the lower case Greek letter rho), although the Latin letter D can also be used. Mathematically, density is defined as mass divided by volume:
ρ
=
m
V
{\displaystyle \rho ={\frac {m}{V}}}
where ρ is the density, m is the mass, and V is the volume. In some cases (for instance, in the United States oil and gas industry), density is loosely defined as its weight per unit volume, although this is scientifically inaccurate – this quantity is more specifically called specific weight.
For a pure substance the density has the same numerical value as its mass concentration.
Different materials usually have different densities, and density may be relevant to buoyancy, purity and packaging. Osmium and iridium are the densest known elements at standard conditions for temperature and pressure.
To simplify comparisons of density across different systems of units, it is sometimes replaced by the dimensionless quantity "relative density" or "specific gravity", i.e. the ratio of the density of the material to that of a standard material, usually water. Thus a relative density less than one relative to water means that the substance floats in water.
The density of a material varies with temperature and pressure. This variation is typically small for solids and liquids but much greater for gases. Increasing the pressure on an object decreases the volume of the object and thus increases its density. Increasing the temperature of a substance (with a few exceptions) decreases its density by increasing its volume. In most materials, heating the bottom of a fluid results in convection of the heat from the bottom to the top, due to the decrease in the density of the heated fluid. This causes it to rise relative to more dense unheated material.
The reciprocal of the density of a substance is occasionally called its specific volume, a term sometimes used in thermodynamics. Density is an intensive property in that increasing the amount of a substance does not increase its density; rather it increases its mass.
Summary: A 5.0- cm -diameter cylinder floats in water. How much work must be done to push the cylinder 11 cm deeper into the water?
F =Aρgx
A 5.0- cm -diameter cylinder floats in water. How much work must be done to push the cylinder 11 cm deeper into the water?
F =Aρgx
x being the...
I feel like I'm missing something fundamental here. Given only the lengths and the densities, how am I supposed to find a numerical centre of mass?
Thought process so far:
Are we supposed to use the ratio of the densities to find this answer? like ##\frac{8g/cm^3}{2.7g/cm^3}##? and then use...
Hi everyone!
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I recently encountered this problem on a test where the solution for the above problem was given as follows:
$$F= \frac{Gm_1m_2} {r^2} $$ (1)
but
$$ m=\frac{4}{3}\pi R^3 $$
substituting in equation (1)
$$F= \frac{{G(\frac{4}{3}\pi R^3\rho})^2 }{2R^2} $$
where r=radii of the two spheres
m=mass...
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http://hyperphysics.phy-astr.gsu.edu/hbase/pbuoy.html#buoy
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Reasoning :
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I think the right choice is c. I'll pass on my reasoning to you:
We can think that if the formula of the potential is
V(r)=\dfrac{kq}{r}
If r tends to infinity, then V(r)=0.
But the correct answer is d).
Yes, pressure is same along the line for both the tubes since the liquids are in equlibrium. (I ignore the slightly longer air column in the left compared to the right).
Is my answer correct?
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We understand that the crucial thing about the problem is that the volume of water present in the three containers are not the same. Also, we note that in each case the weight of the container is the total weight of its contents. (A student might be confused as to why should be so - after all...
https://www.wired.com/story/a-bizarre-form-of-water-may-exist-all-over-the-universe/
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I knew the Black Ice Theories since around 1990
https://www.nature.com/articles/s41586-019-1114-6
-- Demontis, P., LeSar, R. & Klein, M. L. New high-pressure phases of ice. Phys. Rev. Lett. 60...
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We want to calculate the ao/a(teq) of the equilibrium point between ρm and ρr (ρm= ρr )
My book solves it this way;
ρm(t) / ρr(t)= a(t) ⇒
⇒ (ρm/ ρr)teq =1 =
= (ρm/ ρr)o * a(teq)/ ao
I don't understand the a(teq)/ ao part. If ρm(t)= ρο/αo3 and ρr(t)= ρο/αo4 then it should be
ρm(t)/ ρr(t) =...
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Hello,
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MODERATOR'S NOTE: HOMEWORK INCORRECTLY POSTED TO CLASSICAL FORUM, SO NO TEMPLATE
I need help with the following question:
Please have a look at the question and my attempt at the solution.
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Homework Statement
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Homework Equations
I think this may be involved.
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The Attempt at a Solution [/B]
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https://imgur.com/a/wOSD30D
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Good day all!
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Homework Statement
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