An electron initially has a speed 16 km/s along the x-direction and enters an electric field of strength 27 mv/m that points in the y-direction and has a width of 22.5 cm. (a) how long does it take the electron to pass through the field? μs (b) how far is the deflection?

a)

b)S= 46.4 cm

Explanation:

Given that

Velocity = 16 Km/s

V= 16,000 m/s

E= 27 mV/m

E=0.027 V/m

d= 22.5 cm

d= 0.225 m

a)

lets time taken by electron is t

d = V x t

0.225 = 16,000 t

b)

We know that

F = m a = E q                    ------------1

Mass of electron ,m

Charge on electron

So now by putting the values in equation 1

Here initial velocity u= 0 m/s

S=0.464 m

S= 46.4 cm

S is the deflection of electron.

your answer is, orbit around the nucleus

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s