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deff fn [24]
3 years ago
12

Cuál es el método que utiliza el factor de la gravedad y la difusión de un líquido sobre un sólido para separar las partes de un

a mezcla de varios líquidos mezclados entre sí
Physics
1 answer:
Airida [17]3 years ago
4 0

Answer:

chromatography

Explanation:

Chromatography is one of the powerful technique that is used to separate the mixed components of liquid or the gas mixtures.

In this technique the separation of the mixture by passing the mixture in some solution or some suspension through the medium where the components are made to move at some different rates.

The basic idea in this experiment is that the mixture sample is in mobile phase and it is forced either by pumping or by  gravity or by capillary action through the stationary phase to separate the mixture.

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PLEASE HELP I NEED TO TURN IT IN IN AN HOUR ITLL GIVE YOU POINTS PLS PLEASE
miv72 [106K]

Answer:

  1. 17.95025
  2. 172.3995

Explanation:

1.

\mathrm{The\:solution\:for\:Long\:Division\:of}\:\frac{2.995}{0.16685}\:\\\mathrm{is}\:17.95025

2.

\mathrm{Multiply\:without\:the\:decimal\:points,\:then\:put\:the\:decimal\:point\:in\:the\:answer}\\\\910\times\:18945=17239950\\\\910\mathrm{\:has\:}0\mathrm{\:decimal\:places}\\0.18945\mathrm{\:has\:}5\mathrm{\:decimal\:places}\\\\\mathrm{Therefore,\:the\:answer\:has\:}5\mathrm{\:decimal\:places}\\\\=172.39950\\\\Refine\\=172.3995

5 0
3 years ago
Which of the following statements about a metal wire in equilibrium are true? Select all that apply. There cannot be excess char
olga_2 [115]

Answer:

There may be excess charges in the interior of the wire

The net electric field everywhere inside the wire is zero

The interior of the metal wire is neutral.

There may be excess charges on the surface of the wire.

There is no net flow of mobile electrons inside the wire.

Explanation:

For any metal wire in equilibrium position, there may be excess charges in the interior of the wire and the net electric field everywhere inside the wire is zero. Additionally, the interior of the metal wire is always neutral and there is likely to be excess charges on the surface of the wire. Moreover, it's important to note that for a metal wire in equilibrium, there is no net flow of mobile electrons inside the wire.

3 0
4 years ago
A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit
Amanda [17]

A) +1.67 rad/s^2

The angular acceleration of the wheel is given by

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

where

\omega_i = -6.00 rad/s is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

\omega_f = 4.00 rad/s is the final angular velocity (anticlockwise, so with a positive sign)

\Delta t= 6.00 s - 0=6.00 s is the time interval

Substituting into the equation, we find the angular acceleration:

\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant t_1 where the angular velocity becomes positive (so, \omega_i=0) and the time corresponding to the final instant t_2 = 6.0 s, where \omega_f = +6.00 rad/s. We can find this time interval by using

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

And solving for \Delta t we find

\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant t_1=0 when \omega_i=-4.00 rad/s) and the time corresponding to the instant in which the velovity becomes positive t_2, when \omega_f = 0 rad/s. We can find this time interval by using

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

And solving for \Delta t we find

\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{0-(-4.00 rad/s)}{+1.67 rad/s^2}=2.4 s

D) 5.6 rad

The angular displacement of the wheel is given by the equation

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where we have

\omega_i = -6.00 rad/s is the initial angular velocity of the wheel

\omega_f = 4.00 rad/s is the final angular velocity

\alpha=+1.67 rad/s^2 is the angular acceleration

Solving for \theta,

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{((+6.00 rad/s)^2-(-4.00 rad/s)^2}{2(+1.67 rad/s^2)}=5.6 rad

3 0
3 years ago
PRELIMINARY QUESTIONS
zimovet [89]
French fries were on your car
3 0
3 years ago
A thermostat is used for which of the following? maintaining the temperature in a room turning a heating system on turning a hea
Molodets [167]
All of the above, it's a very useful device :P 
4 0
3 years ago
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