W=q(Delta V) = Change in kinetic energy, which is (-1.6*10^-19 C)(3000V) = -4.8*10^-16 N*m
-4.8*10^-16 N*m = .5*(9.109*10^-31 kg)*(v^2)
Since a newton is kg*m/s^2, the kg's cancel and im left with m^2/s^2
Sorry for the confusion.
I need to find the force. Once I find that everything else isn't difficult.
Find the speed of the electron:
W=q(Delta V) = F*d, where W is -4.8*10^-16 N*m
Find the distance between the been entering and exiting the field.
r=mv/(eB) where r is the radius so...
An electron gun (applied voltage of 3000 volts) is emitting electrons. THe electrons enter a region of constant magnetic field (B=.025 Tesla.) The magnetic field is perpendicular to the velocity of the beam.
W=Q(Delta V) = F*d
I was given a question about a certain material. The idea was that when it was heated from zero degrees C to 200 degrees C, it would be 10.06 mm from its original length of 10. Doing it in reverse order, length of 10.06 mm from 200 C to 0 C gave me an answer of 9.99964, which is less than the...