1.5 / 0.5 = 3 I believe this is the right answer
.Answer;
Using Fmax=qVB
F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)
ANS=1.29*10^-12 N
2. Using Amax=Fmax/ m
Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)
ANS=1.93*10^15 m/s^2*
3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton
It just completely stops kinetic energy is movement so technically it would stop.
Explanation:
Let omega = angular velocity (in rad/s). Then
omega = (# of oscillations)/(6 s)
= (30 osc)/(6 s) = 5 osc/s
We need to convert this to rad/s:
omega = (5 osc/s)(2π rad/osc)
= 10π rad/s
= 31.4 rad/s
The measurement of the wave trough
