The effort is always less than the load for a second-class lever. true or false
2 answers:
You might be interested in
Answer:
Given that
speed u=4*10^6 m/s
electric field E=4*10^3 N/c
distance b/w the plates d=2 cm
basing on the concept of the electrostatices
now we find the acceleration b/w the plates to find the horizontal distance traveled by the electron when it hits the plate.
acceleration a=qE/m==
m/s
now we find the horizontal distance traveled by electrons hit the plates
horizontal distance
=
== 3 cm