Physics, 21.02.2020 17:58 Yeajay6313

Explain the changes in molecular motion of particles as they change from solid to liquid to gas. A) As the phase changes occur, the freedom of motion of the particles increases. B) As the phase changes occur, the freedom of motion of the particles decreases. C) As the phase changes occur, the freedom of motion of the particles compresses. D) As the phase changes occur, the freedom of motion of the particles is constant.

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Answer from: ajajjsisi9x8

The correct statement is option A : As the phase changes occur, the freedom of motion of the particles increases.

At the solid state, the molecules of solid are tightly packed and they are unable to move freely. When heat is applied to solid, it melts and the molecules of the solid gains kinetic energy and therefore tends to shift away due to heat.

This kinetic energy possesses by this molecules allow them to move freely. When further heat is applied to the liquid above the boiling point the molecules of the water vapourises and these molecules gain more kinetic energy more than they do in liquid state and were able to move freely more than in liquid.

This shows that the freedom of the particles (molecules) increase with change in state from solid to gas.

Therefore we conclude that as the phase change occurs, the freedom of motion of of the particles increases.

Hence The correct statement is option A : As the phase changes occur, the freedom of motion of the particles increases.

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Answer from: Jasten

As the phase changes occur, the freedom of motion of the particles increases

Explanation:

As the phase changes occur from solid to gas, the freedom of motion of the particles increases

At the solid state (let's say ice), the molecules of the solid are tightly packed and they are unable to move freely. When heat is applied to the state of the solid, ice it melts and the molecules of the solid gains kinetic energy and therefore tends to shift away from their mean fixed position due to heat. This kinetic energy possesses by this molecules allow them to move freely (energy due to motion). When further heat is applied to the liquid (above the boiling point), the molecules of the water vapourizes(turns to gas) and this molecules gain more kinetic energy more than they do in liquid state and were able to move freely more than in liquid. This shows that the freedom of the particles (molecules) increase with change in state from solid to gas.

Answer from: Quest

Electoral current force mechanical advantage if the wheel and axle is 2 n and the output force is 20 n

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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