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3 years ago

What are some techniques for separating mixtures into their components?

Physics

1 answer:

andreev551 [17]3 years ago

3 0

Some commonly used techniques might be distillation and magnetism.

Distillation is basically a method of separating liquid mixture by basically boiling and condensing the different liquids. Since liquids don't have the same boiling points. one will evaporate first, and the other liquid or liquds will remaim.

Magnetism is a method of separating solid mixtures. One or some solids will get atracted to the magnet and the others will remain. Note that this only works if one of the solids has a magnetical property.

Hope it helped,

Happy homework/ study/ exam!

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A solid sphere has a temperature of 614 K. The sphere is melted down and recast into a cube that has the same emissivity and emi

krok68 [10]

Answer:581.87 K

Explanation:

Given

Sphere is melted to form a square

Let the radius of sphere be r and square has a side a

Therefore

$\frac{4\pi}{3}r^3=a^3$

Surface area of sphere $A_s=4\pi r^2$

Surface area of cube $A_c=6a^2$

Total emmisive remains same

Thus $P=A\epsilon \sigma T^4$

$A_sT_s^4=A_cT_c^4$

$\frac{T_c^4}{T_s^4}=\frac{A_s}{A_c}$

$\frac{T_c^4}{T_s^4}=\frac{1}{2}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{3}}$

$\frac{T_c}{T_s}=\frac{1}{2^{0.25}}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{12}}$

$T_c=T_s\times \frac{1}{2^{0.25}}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{12}}$

$T_c=614\times \frac{1.12679}{1.189}$

$T_c=581.87 K$

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3 years ago

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stira [4]

Answer:

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3 years ago

Place /our live /music / important / has / in / an

oksian1 [2.3K]

Answer:

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4 years ago

A coin and feather are dropped in a moon. what will fall earlier on ground.give reasons.if they are dropped in the earth,which o

Sergeeva-Olga [200]

Answer:

on the moon, they will fall at the time

on earth, the coin will fall faster to the ground

Explanation:

A coin and feather dropped in a moon experience the same acceleration due to gravity as small as 1.625 m/s², and because of the absence of air resistance both will fall at the same rate to the ground.

If the same coin and feather are dropped in the earth, they will experience the same acceleration due to gravity of 9.81 m/s² and because of the presence of air resistance, the heavier object (coin) will be pulled faster to the ground by gravity than the lighter object (feather).

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3 years ago

A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant k=450N/m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
 $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
 $E=K_{max} = \frac{1}{2}mv_{max}^2$
Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
 $E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
 $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
 $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b) the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
 $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since
 $E=U+K$
we find the kinetic energy when the glider is at this position:
 $K=E-U=0.36 J - 0.05 J = 0.31 J$
And then we can find the corresponding velocity:
 $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
 $a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
 $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$

e) the total mechanical energy of the glider at any point in its motion. 

we have already calculated it at point b), and it is given by
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

4 years ago