The kinetic energy of a gas molecule is given by the equation: \[ E = \frac{3}{2} k_B T \] where \( E \) is the kinetic energy, \( k_B \) is the Boltzmann constant, and \( T \) is the temperature. For two gases at the same temperature, the kinetic energy depends only on the temperature, which is the same for both gases.
Therefore, the kinetic energies of the gases are equal: \[ E_1 = E_2 \]
Thus, the correct answer is \( E_1 = E_2 \).
The motion of a particle in the XY plane is given by \( x(t) = 25 + 6t^2 \, \text{m} \); \( y(t) = -50 - 20t + 8t^2 \, \text{m} \). The magnitude of the initial velocity of the particle, \( v_0 \), is given by:
Consider a rope fixed at both ends under tension so that it is horizontal (i.e. assume the rope is along x-axis, with gravity acting along z-axis). Now the right end is continually oscillated at high frequency n (say n=100 Hz) horizontally and in a direction along the rope; amplitude of oscillation is negligible. The oscillation travells along the rope and is reflected at the left end.
Let the total length of rope be l, total mass be m and the acceleration due to gravity be g.
After initial phase (say a mintue or so), the rope has __(BLANK-1)__ wave, which is __(BLANK-2)__ in nature. It results from superposition of left travelling and right travelling __(BLANK-3)__ waves. This resulting wave has a frequency __ (BLANK-4)_ that of oscillation frequency nu. Simple dimensional analysis indicates that the frequency of can be of the form: ___(BLANK-5)__ .
Two point charges M and N having charges +q and -q respectively are placed at a distance apart. Force acting between them is F. If 30% of charge of N is transferred to M, then the force between the charges becomes:
If the ratio of lengths, radii and Young's Moduli of steel and brass wires in the figure are $ a $, $ b $, and $ c $ respectively, then the corresponding ratio of increase in their lengths would be: