`M_1` and `M_2` represent the masses of `_(10)Ne^(20)` nucleus and `_(20)Ca^(40)` nucleus respectively.State ,whether `M_1=2 M_1 or M_2>2 M_1 ` or `M_2ltM_1 ?`

Two perfect gases at absolute temperatures T_1 and T_2 are mixed. There is no loss of energy. Find the temperature of the mixture if the masses of the molecules are m_1 and m_2 and the number of the molecules in the gases are n_1 and n_2 respectively.

m_1 and m_2 are the masses of two bodies. When a force of 5 N is applied on each body, m_1 gets an acceleration of 10 m//s^2 and m_2, 20 m//s^2. If the two bodies are tied together and the same force is applied, find the acceleration of the combined system.

Supposing Newton's law of gravitational for gravitation fores vecF_1 and vecF_2 between two masses m_1 and m_2 at positions vecr_1 and vecr_2 read vecF_1 = - vecF_2 = - (vecr_12)/(r_12^3) GM_0^1((m_1m_2)/(m_0^2))^n where M_0 is a constant of dimension of mass. vecr_12 = vecr_1 - vecr_2 and n is a number. In such a case

Calculate the binding energy per nucleon of "_20Ca^40 nucleus. Given mass of "_20Ca^40 nucleus = 39.962589 a.m.u., mass of proton = 1.007825 a.m.u., mass of neutron = 1.008665 a.m.u. and 1 a.m.u. = 931.5 MeV.

If a gas contains molecules of two different masses m_1 and m_2 both at the same temperature, then (v_1/ v_2)_(r.m.s.) is: