A proton starts from rest, and moves 1 meter through a constant electric field. The potential difference between its starting and ending points is 500 V. How fast is the proton moving at the end? How large is the electric field? How long does it take the proton to complete its journey? [Late clarification: since the field is accelerating the particle from rest, you can assume that the direction of the particle's motion is parallel to the field.]

E = - 500 V / m,  v = 30.95 10⁴ m / s   and    t = 6.46 10⁻⁶ s

Explanation:

For this problem we use the relation

           ΔU = - E s

           E = -ΔU / s

           E = - 500/1

           E = - 500 V / m

Now we can look for the proton approach

           F = q E

let's use Newton's second law

           F = m a

           a = F / m

           a = q E / m

Now let's use kinematics relations, where the proton starts from the rest

v₀ = 0

           v² = v₀² + 2 a x

           v = √2 q E / m x

           v = √ (2 1.6 10⁻¹⁹ 500 / 1.67 10⁻²⁷  1)

           v = √ (958.08 108)

           v = 30.95 10⁴ m / s

for time let's use the equation

          x = v₀ t + ½ to t2

          t = √2x / a

          t = √ (2x m / qE)

          t = √ (2 1 1.67 10⁻²⁷ / (1.6 10⁻¹⁹ 500))

          t = √ (0.004175 10⁻⁸)

          t = 0.0646 10⁻⁴ s

          t = 6.46 10⁻⁶ s

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge

answer: the metamorphic rocks are formed when they are exposed to high pressure and high temperatures. this happens when one of the earth's plates comes in contact with the other plate by traveling in opposite directions, straight into each other or slipping under one another. the friction that is caused by this activity results in   melting of rocks and when it cools they are known as metamorphic rocks. this process needs a controlled amount of pressure and temperature otherwise it may turn into magma.