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3 years ago

15

Rock salt contains sand and salt. A student mixed rock salt in warm water. Describe how the student would separate out the sand,

salt and water from the mixture formed

1 answer:

geniusboy [140]3 years ago

4 0

Answer:

Explanation:

First, he can add the rock salt to the water and mix it well. Then he can remove the sand by decantation, after the sand has sedimented. He can seperate the water from the salt by evaporation. He should evaporate the water completely and capture the steam. the steam will cool into water. After evaporating completely, the salt will remain in the container

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nika2105 [10]

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3 years ago

When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart

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Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

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3 years ago

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3 years ago

enot [183]

Answer:

A. $U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. $U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. $U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

$C=\dfrac{\epsilon A}{d}$

$C$ is the capacitance, $A$ is the common plate area, $d$ is the plate separation and $\epsilon$ is the permittivity of the material between the plates.

For air or free space, $\epsilon$ is $\epsilon_0$ called the permittivity of free space. In general, $\epsilon=\epsilon_r \epsilon_0$ where $\epsilon_r$ is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, $\epsilon_r=1$ .

The energy stored in a capacitor is the average of the product of its charge and voltage.

$U = \dfrac{QV}{2}$

Its charge, $Q$ , is related to its capacitance by $Q=CV$ (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for $U$ ,

$U = \dfrac{CV^2}{2}$

A. Substituting for $C$ in $U$ ,

$U_0 = \dfrac{\epsilon_0 A V^2}{2d}$

B. When the distance is $3d$ ,

$U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}$

$U_1 = \dfrac{\epsilon_0 A V^2}{6d}$

C. When the distance is restored but with a dielectric material of dielectric constant, $K$ , inserted, we have

$U_2 = \dfrac{K\epsilon_0 A V^2}{2d}$

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AveGali [126]

Answer:

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