The first one. The E and B chatacteristic are perpendicular to eachother. The direction of the wave can be found by the right hand rule.
Answer:
ni = 2.04e19
Explanation:
we know that in semiconductor like intrinsic, when electron leave the band, it leave a hole in valence band so we have
n = p = ni
from intrinsic carrier concentration



1.7 = ni * 1.6*10^{-19} * (.35 + .17)
ni = 2.014 *10^{19} m^{-3}
ni = 2.04e19
Answer:
The strength of the magnetic field is
.
Explanation:
Given that,
Length of the rod, L = 1.01 m
Speed with which the rod is moving, v = 3.47 m/s
We need to find the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.265 mV across the rod. When the rod is moving with some speed, an emf gets induced and it is given by :

B is magnetic field

So, the strength of the magnetic field is
.
Answer:
v = 3.27 m/s
Explanation:
KE = 1/2 mv^2
695 J = 1/2 (130kg)(v^2)
695 J / (1/2 x 130kg) = v^2
v^2 = square root of 10.69
v = 3.27 m/s