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

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

air

in a tire you fill a tire with air

then a couple days later you check the air pressure in the tire

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