Arelativistic proton has a momentum of 1.0 × 10-19 kg · m/s and a rest energy of 0.15 nj. what is the kinetic energy of this proton? (c = 3.00 × 108 m/s, mproton = 1.67 × 10-27 kg)

The kinetic energy is 0 J.

Explanation:

The kinetic energy of a relativistic proton is determined as follows:

KE = mc² - m₀c²

To determine m₀, we use the expression

E_rest = m₀c²

where

Thus, rearranging the above expression we get

m₀ = E_rest/c²

    =(0.15×10⁻⁹)/(3×10⁸)²

    = 1.67×10⁻²⁷ kg

Therefore,

KE = (1.67×10⁻²⁷)(3×10⁸)² - (1.67×10⁻²⁷)(3×10⁸)²

     = 0 J

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

the answer to your question is c.

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