Question

Определите скорость частицы при которой ее энергия больше ее энергии покоя в k=4 раза?

Answer 1

Вроде бы три но я сомневаюсь но наверняка да да наверняка

Answer 2

Determining the Speed of a Particle with Energy Greater than its Rest Energy by a Factor of k=4

To determine the speed of a particle at which its energy is greater than its rest energy by a factor of k=4, we can use the equation that relates energy and velocity for a particle with mass m:

E = γmc^2

Where: - E is the total energy of the particle, - γ is the Lorentz factor, given by γ = 1/√(1 - v^2/c^2), - m is the rest mass of the particle, - c is the speed of light in a vacuum, approximately 3 x 10^8 meters per second, - v is the velocity of the particle.

We want to find the velocity v at which the energy E is k=4 times greater than the rest energy mc^2. Therefore, we can set up the following equation:

E = kmc^2

Substituting the expression for E in terms of γ and mc^2, we have:

γmc^2 = kmc^2

Canceling out the mc^2 terms, we get:

γ = k

Now, we can solve for the velocity v by substituting γ = 1/√(1 - v^2/c^2) and k = 4 into the equation:

1/√(1 - v^2/c^2) = 4

To solve this equation, we can square both sides:

1/(1 - v^2/c^2) = 16

Cross-multiplying, we have:

1 = 16(1 - v^2/c^2)

Expanding and rearranging the equation, we get:

v^2/c^2 = 1 - 1/16

Simplifying further:

v^2/c^2 = 15/16

Taking the square root of both sides:

v/c = √(15/16)

Finally, solving for v:

v = c√(15/16)

Now, we can calculate the value of v:

v = c√(15/16) ≈ 0.968c

Therefore, the speed of the particle at which its energy is greater than its rest energy by a factor of k=4 is approximately 0.968 times the speed of light.