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zysi [14]
3 years ago
7

Describe some ways that industry and agriculture use physical properties to separate substances.

Physics
2 answers:
musickatia [10]3 years ago
3 0
This is more chemistry. But it is a process called fractional distillation, and it basically separates the long chained hydrocarbons from the short chained hydrocarbons through separation dependant on the boiling point of the crude oil.
Oliga [24]3 years ago
3 0

Some ways that industry and agriculture use physical properties to separate substances are as follows-  

a) Separation based on the difference in DENSITY – In the process of sedimentation, the heavier particle settles at the bottom while the lighter particle remains dissolved or float in the solvent. The settled heavier substance is then removed periodically. For example mud and sand impurities from water are removed through this process.

b) Separation based on the difference in SOLUBILITY – Some substances that remain un dissolved in any other substance can be easily removed through the process of filtration. For example –  Extraction of coffee from grounds

c) Separation based on the difference in Special metal Property – Iron possesses magnetism and thus it can be easily separated from nonmagnetic substances

d) Separation based on the difference in  Vapour pressure/Boiling Point – It is the concept behind the process of separation through Distillation. In distillation,  liquid with the lower boiling point boils first which is collected through condensation while the liquid with higher boiling point remain in the flask.  

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How should the distance, D, between two point charges, q1 and q2, be changed to double their electric potential energy
Licemer1 [7]

The electrostatic potential energy, U, of one point charge q at position d in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force to bring it from the reference position d to that position

u \:  =  \:  \frac{kq1q2}{d}

Thus, to double the electric potential energy U we need to reduce the distance of separation by half (1/2) because they are inversely proportion

6 0
3 years ago
All biomes don’t have the same level of biodiversity. What seems to be the optimal conditions for high biodiversity?
irinina [24]

Answer:

See the answer below

Explanation:

The optimal conditions for high biodiversity seem to be a <u>warm temperature</u> and <u>wet climates</u>.

<em>The tropical areas of the world have the highest biodiversity and are characterized by an average annual temperature of above 18 </em>^oC<em> and annual precipitation of 262 cm. The areas are referred to as the world's biodiversity hotspots. </em>

Consequently, it follows logically that the optimal conditions for high biodiversity would be a warm temperature of above 18 ^oC and wet environment with annual precipitation of not less than 262 cm.

The variation in temperature and precipitation across biomes can thus be said to be responsible for the variation in the level of biodiversity in them.

6 0
3 years ago
In your daily life you come across a range of motion in which acceleration is in the direction of motion
irakobra [83]
That makes no sense to me somehow
3 0
3 years ago
What happens during the process of deposition
frez [133]
Deposition is the process in which sediments, soil and rocks are added to a landform or landmass. When previous weathers surface material , is deposited to a building layer of sediment .
8 0
3 years ago
A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t
sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, \rho=2.44\times 10^{-8}\ \Omega-m

Resistance in terms of resistivity is given by :

R=\dfrac{\rho l}{A}

Also, V = IR

So,

\dfrac{V}{I}=\dfrac{\rho l}{A}

A is area of wire,

\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}, r is radius, r = d/2 (diameter=d)

\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m

Out of four option, near option is (C) 17 μm.

6 0
3 years ago
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