A train 90m long stops in a train station with its front buffers in line with a lamp post on the platform. Later it starts off with an average acceleration of 0.45m/s. What will be its speed in kilometers per hour when the tail buffers pass the lamp post P?​

The speed of the train when the tail buffers pass the lamp post is 9 m/s

Explanation:

The given parameters are;

The length of the train = 90 m

The initial velocity of the train, u = 0

The average acceleration of the train, a = 0.45 m/s²

When the tail buffer pass the lamp post, P, we have;

The distance travelled, s = The length of the train = 90 m

The kinematic equation of motion is given as follows;

v² = u² + 2·a·s

Where;

v = The speed of the train when the tail buffers pass the lamp post

∴ v² = 0² + 2 × 0.45 m/s² × 90 m = 81 m²/s²

Therefore;

v = √(81 m²/s²) = 9 m/s

The speed of the train when the tail buffers pass the lamp post, v = 9 m/s.

beta is electron from nucleus

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

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..