Physics, 30.12.2019 09:31, graceaowen

What fraction of a piece of lead will be submerged when it floats in mercury? the density of lead is 11.3×103kg/m3 and density of mercury is 13.6×103kg/m3?

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answered: j015

According to the Archimedes’ Principle, a body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.

In this case we have a piece of lead floating (partially immersed) in liquid mercury. If we observe in the image attached, this object does not completely fall to the bottom because the net force acting on it is zero, this means it is in equilibrium.

This is due to Newton’s first law of motion, that estates if a body is in equilibrium the sum of all the forces acting on it is equal to zero:

$F_{total}=0$ (1)

Here, the piece of lead experiences two kind of forces:

A force downwards because of its weight:

W=mg (2)

Where is the mass of the piece of lead and the acceleration of gravity, which is $9.8\frac{m}{s^{2}}$

And a force applied upwards because of the buoyancy:

$F_{buoyancy}=d_{m}gV_{i}$ (3)

Where;

$d_{m}$ is the density of liquid mercury which is $13.6(10^{3})\frac{kg}{m^{3}}$

And $V_{i}$ is the volume of the immersed part of the piece of lead

Well, if the total net force exerted to the object is zero:

$F_{total}= F_{buoyancy}-W=0$

Note has a negative sign because this force is applied downwards

This means the buoyancy is equal to the weight:

$F_{buoyancy}=W$ (4)

$d_{m}gV_{i}=mg$ (5)

Dividing by in both sides:

$d_{m}V_{i}=m$ (6)

Now, the density of an object or fluid is the relation between the mass and the volume, in the case of the object of this problem (the piece of lead) is:

$d=\frac{m}{V}$ (7)

Where is the Volume of the piece of lead.

From (7) we can find the mass of the piece of lead:

m=dV (8)

and substitute this value in (6), in order to get a relation between the density of the object and the density of the liquid mercury (the fluid), as follows:

$d_{m}V_{i}=dV$

$\frac{V_{i}}{V}=\frac{d}{d_{m}}$ (9)

$\frac{d}{d_{m}}=\frac{11.3(10^{3})kg/m^{3}}{13.6(10^{3})kg/m^{3}}$

Then;

$\frac{11.3(10^{3})kg/m^{3}}{13.6(10^{3})kg/m^{3}}=0.83$

This result is the same as $\frac{83}{100}$ which represents the $83\%$

Therefore:

$83\%$ of the mass of the piece of lead is immersed in mercury

$What fraction of a piece of lead will be submerged when it floats in mercury? the density of lead i$

answered: Guest

An atom (ours, not carroll's) consists of a positively charged nucleus, surrounded by one or more negatively charged particles called electrons. the positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral. most of an atom's mass is in its nucleus; the mass of an electron is only 1/1836 the mass of the lightest nucleus, that of hydrogen. although the nucleus is heavy, it is quite small compared with the overall size of an atom. the radius of a typical atom is around 0.1 to 0.25 nanometres (nm), whereas the radius of a nucleus is about 10-6 nm. * if an atom were enlarged to the size of the earth, its nucleus would be only 200 feet in diameter and could easily rest inside a small football stadium.read more on -

answered: Guest

If there was no gravity then their would be no downword force on the football to have it return the the grownd, and if there was no friction (mainly air friction) then there would be no opposing force the stop the ball from moving through the air allowing the ball to travle ∞.

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What fraction of a piece of lead will be submerged when it floats in mercury? the density of lead i...