The temperature at which all molecular motion stops is

Physics

1 answer:

34kurt3 years ago

7 0

Answer:

Zero Kelvin

Explanation:

The average kinetic energy of the particles in a gas is related to the absolute temperature of the gas by (for an ideal monoatomic gas):

$E_K = \frac{3}{2}kT$

where

k is the Boltzmann constant

T is the absolute temperature

The average kinetic energy is the energy possessed by the particles due to their motion; we see that this energy becomes zero when T = 0, which means when the substance reaches a temperature of zero Kelvin. Therefore, this means that at this temperature all the particles stop moving.

You might be interested in

Describe how a change in resistance would affect the current in a circuit.

Tresset [83]

Explanation:

The relation between potential difference, current and resistance flowing in a circuit is given by using Ohm's law. It can be given by :

V = IR

$I=\dfrac{V}{R}$

Resistance opposes the flow of electric current in the circuit. It means that, if resistance is more, less current will flow through the circuit.

3 0

3 years ago

A negative charge is at rest at the origin of an axis system. Location x is at coordinate point (2m,3m) while location y is at (

Sergeu [11.5K]

The magnitude of the E-field decreases as the square of the distance from the charge, just like gravity.

Location ' x ' is √(2² + 3²) = √13 m from the charge.

Location ' y ' is √ [ (-3)² + (-2)² ] = √13 m from the charge.

The magnitude of the E-field is the same at both locations.

The direction is also the same at both locations ... it points toward the origin.

5 0

3 years ago

A rock has a specific gravity of 2.32 and a volume of 8.64 in3 how much does it weigh

lianna [129]

Answer: 3.21 N

$Specific\hspace{1mm} gravity = \frac {Density\hspace{1mm}of\hspace{1mm}substance}{Density\hspace{1mm} of \hspace{1mm}water}$

$\Rightarrow Density \hspace{1mm}of\hspace{1mm}substance= 2.32\times 1000\hspace{1mm} kg/m^3 = 2320\hspace{1mm}kg/m^3\\ Mass =Density\times volume\\ \Rightarrow 2320 \hspace{1mm} kg/m^3\times 8.64 \hspace{1mm}in^3\times \frac {1.64\times10^{-5} m^3}{1\hspace{1mm}in^3}=0.328 kg$