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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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Answer: 6s

Explanation:

Vs=32m/s  speed at beginning of slowing down

Vf=0m/s     stop speed

a= -6 m/s²  acceleration

----------------

Use equation for acceleration :

a=(Vf-Vs)/t

a*t=Vf-Vs

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t=(0-36)/-6

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The flagpole is held vertical by two ropes. The first of these ropes has a tension in it of 100 N and is at an angle of 60° to t
KatRina [158]

Answer:

T₂ = 123.9 N,  θ = 66.2º

Explanation:

To solve this exercise we use the law of equilibrium, since the diaphragm does not appear, let's use the adjoint to see the forces in the system.

The tension T1 = 100 N, we create a reference frame centered on the pole

X axis

       T₁ₓ - T_{2x} = 0

        T_{2x}= T₁ₓ

Y axis y

      T_{1y} + T_{2y} - 200N = 0

      T_{2y} = 200 -T_{1y}

let's use trigonometry to find the component of the stresses

         sin 60 = T_{1y} / T₁

         cos 60 = t₁ₓ / T₁

         T_{1y} = T₁ sin 60

         T1x = T₁ cos 60

         T_{1y}y = 100 sin 60 = 86.6 N

         T₁ₓ = 100 cos 60 = 50 N

for voltage 2 it is done in the same way

         T_{2y} = T₂ sin θ

         T₂ₓ = T₂ cos θ

we substitute

         

           T₂ sin θ= 200 - 86.6 = 113.4

           T₂ cos θ = 50              (1)

to solve the system we divide the two equations

           tan θ = 113.4 / 50

           θ = tan⁻¹ 2,268

           θ = 66.2º

we caption in equation 1

           T₂ cos 66.2 = 50

           T₂ = 50 / cos 66.2

           T₂ = 123.9 N

8 0
3 years ago
A flute player hears four beats per second when she compares her note to a 523 HzHz tuning fork (the note C). She can match the
laiz [17]

Answer:

527 Hz

Solution:

As per the question:

Beat frequency of the player, \Delta f = 4\ beats/s

Frequency of the tuning fork, f = 523 Hz

Now,

The initial frequency can be calculated as:

\Delta f = f - f_{i}

f_{i} = f \pm \Delta f

when

f_{i} = f + \Delta f = 523 + 4 = 527 Hz

when

f_{i} = f - \Delta f = 523 - 4 = 519 Hz

But we know that as the length of the flute increases the frequency decreases

Hence, the initial frequency must be 527 Hz

7 0
3 years ago
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Irina-Kira [14]
Density = Mass divided by Volume
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