Which of the following statements is true?

Physics

2 answers:

kupik [55]3 years ago

6 0

Answer: The correct answer is "The freezing and melting temperatures of a substance are the same".

Explanation:

Freezing temperature: It is the temperature at which the liquid becomes the solid.

Melting temperature: It is the temperature at which the solid starts converted into liquid.

The freezing point of a substance is the same as the melting point of the substance. During melting and freezing of the substance, the temperature remains same as the energy is spent in changing the phase of the substance not in changing the temperature.

Latent heat is there. It is a hidden heat. It is required to change phase of the substance without showing temperature on thermometer.

Boiling temperature: It is the temperature at which liquid gets converted into vapors. At this temperature, vapor pressure of the liquid equals to the pressure surrounding the liquid.

xenn [34]3 years ago

4 0

Correct answer choice is :

C) The freezing and melting temperatures of a substance are the same.

Explanation:

Fluids have a particular temperature at which they convert into solids, identified as their freezing point. In theory, the melting point of a solid should be the same as the freezing point of the liquid. In practice, small variations among these measures can be seen. The freezing point of a matter is the same as that substance's melting point. At this distinct temperature, the substance can exist as either a solid or a liquid. At temperatures below the freezing/ melting point, the substance is a solid.

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Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?

storchak [24]

Answer:

$\large \boxed{42\, \mu \text{C}}$$

Explanation:

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The charges are identical, so we can write the formula as

$F=k\dfrac{q^{2}}{r^2}$

$\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}$