A proton gains its mass twice of its rest mass while moving in an energy field.What is the velocity of the proton?The rest mass of the proton is 1.67*10^-27 kg.

When a proton gains its mass to twice its rest mass while moving in an energy field, this indicates relativistic effects. According to Einstein's theory of relativity, the relativistic mass $m$ of a moving object is related to its rest mass $m_0$ by the Lorentz factor $\gamma$:

$m = \gamma m_0$

Given that the moving mass $m$ is twice the rest mass $m_0$:

$m = 2m_0$

We know that:

$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$

Substituting $m = 2m_0$:

$2m_0 = \gamma m_0$

$2 = \gamma$

Therefore:

$\gamma = 2$

Now, solving for $v$ using the definition of $\gamma$:

$2 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$

Squaring both sides:

$4 = \frac{1}{1 - \frac{v^2}{c^2}}$

$1 - \frac{v^2}{c^2} = \frac{1}{4}$

$\frac{v^2}{c^2} = 1 - \frac{1}{4}$

$\frac{v^2}{c^2} = \frac{3}{4}$

$v^2 = \frac{3}{4} c^2$

$v = c \sqrt{\frac{3}{4}}$

$v = c \sqrt{\frac{3}{4}}$

$v = c \cdot \frac{\sqrt{3}}{2}$

Since the speed of light $c$ is $3 \times 10^8$ meters/second:

$v = 3 \times 10^8 \cdot \frac{\sqrt{3}}{2} \, \text{m/s}$

$v \approx 3 \times 10^8 \cdot 0.866 \, \text{m/s}$

$v \approx 2.598 \times 10^8 \, \text{m/s}$