This leads to a paradox known as the Gibbs paradox, after Josiah Willard Gibbs. The paradox allows for the entropy of closed systems to decrease, violating the second law of thermodynamics. A related paradox is the "mixing paradox".
lf a heavy point mass is suspended by a weightless, inextensible and perfectly flexible string from a rigid support, then this arrangement is called simple pendulum.
In practice, however, these requirements cannot be fulfilled. So we use a practical pendulum.
A practical pendulum consists of a small metallic solid sphere suspended by a fine silk thread from a rigid support. This is the practical simple pendulum which is nearest to the ideal simple pendulum.
Note :
The metallic sphere is called the bob.
When the bob is displaced slightly to one side from its mean position and released, it oscillates about its mean position in a vertical plane.
Answer:
![K.E.=\dfrac{3}{2}KT](https://tex.z-dn.net/?f=K.E.%3D%5Cdfrac%7B3%7D%7B2%7DKT)
Explanation:
Given that
Number of particle =N
Equilibrium temperature= T
Side of cube = L
Gravitational acceleration =g
The kinetic energy of an atom given as
![K.E.=\dfrac{3}{2}KT](https://tex.z-dn.net/?f=K.E.%3D%5Cdfrac%7B3%7D%7B2%7DKT)
Where
Equilibrium temperature= T
Boltzmann constant =K
K =1.380649×10−23 J/K
To solve the problem it is necessary to apply the equations related to the conservation of both <em>kinetic of rolling objects</em> and potential energy and the moment of inertia.
The net height from the point where it begins to roll with an inclination of 30 degrees would be
![h=Lsin30](https://tex.z-dn.net/?f=h%3DLsin30)
![h=10sin30](https://tex.z-dn.net/?f=h%3D10sin30)
![h=5m](https://tex.z-dn.net/?f=h%3D5m)
In the case of Inertia would be given by
![I = \frac{mR^2}{2}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BmR%5E2%7D%7B2%7D)
In general, given an object of mass m, an effective radius k can be defined for an axis through its center of mass, with such a value that its moment of inertia is
![I = mk^2](https://tex.z-dn.net/?f=I%20%3D%20mk%5E2)
![\frac{mR^2}{2}= mk^2](https://tex.z-dn.net/?f=%5Cfrac%7BmR%5E2%7D%7B2%7D%3D%20mk%5E2)
![\frac{k^2}{R^2}=\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bk%5E2%7D%7BR%5E2%7D%3D%5Cfrac%7B1%7D%7B2%7D)
Replacing in Energy conservation Equation we have that
Potential Energy = Kinetic Energy of Rolling Object
![mgh = \frac{1}{2}mv^2(1+\frac{k^2}{r^2})](https://tex.z-dn.net/?f=mgh%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%281%2B%5Cfrac%7Bk%5E2%7D%7Br%5E2%7D%29)
![9.8*5=\frac{1}{2}v^2(1+\frac{1}{2})](https://tex.z-dn.net/?f=9.8%2A5%3D%5Cfrac%7B1%7D%7B2%7Dv%5E2%281%2B%5Cfrac%7B1%7D%7B2%7D%29)
![v^2 (1.5) = 98](https://tex.z-dn.net/?f=v%5E2%20%281.5%29%20%3D%2098)
![v=8.0829m/s](https://tex.z-dn.net/?f=v%3D8.0829m%2Fs)
Therefore the correct answer is C.
The correct answer is B. The addition of oxygen will increase the rate of combustion. oxygen and raw material (fuel) are the substrates for the combustion process, therefore, increase in the same will obviously increase combustion rate. Carbon in the fuel is converted into carbon dioxide and heat.