Formula for feild strength= F/q
q=7.0^10-5 coulombs
F=5.2 N
E=5.2 / 7.0^10-5
E=
Answer:
a) A = 3 cm, b) T = 0.4 s, f = 2.5 Hz,
2) A standing wave the displacement of the wave is canceled and only one oscillation remains
Explanation:
a) in an oscillatory movement the amplitude is the highest value of the signal in this case
A = 3 cm
b) the period of oscillation is the time it takes for the wave to repeat itself in this case
T = 0.4 s
the period is the inverse of the frequency
f = 1 /T
f = 1 /, 0.4
f = 2.5 Hz
2) a traveling wave is a wave for which as time increases the displacement increases, in the case of a transverse wave the oscillation is perpendicular to the displacement and in the case of a longitudinal wave the oscillation is in the same direction of the displacement.
A standing wave occurs when a traveling wave bounces off some object and there are two waves, one that travels in one direction and the other that travels in the opposite direction. In this case, the displacement of the wave is canceled and only one oscillation remains.
Answer:
the vertical position is 1.1971m
Explanation:
Recall that
recall that 
this implies that distance = 12 * 5.0
= 60
therefore; 
y = 1.1971m
the
Answer: The unpolarized light's intensity is reduced by the factor of two when it passes through the polaroid and becomes linearly polarized in the plane of the Polaroid. When the polarized light passes through the polaroid with the plane of polarization at an angle 
with respect to the polarization plane of the incoming light, the light's intensity is reduced by the factor of 
(this is the Law of Malus).
Explanation: Let us say we have a beam of unpolarized light of intensity 
that passes through two parallel Polaroid discs with the angle of between their planes of polarization. We are asked to find such that the intensity of the outgoing beam is 
. To solve this we follow the steps below:
Step 1. It is known that when the unpolarized light passes through a polaroid its intensity is reduced by the factor of two, meaning that the intensity of the beam passing through the first polaroid is
This beam also becomes polarized in the plane of the first polaroid.
Step 2. Now the polarized beam hits the surface of the second polaroid whose polarization plane is at an angle with respect to the plane of the polarization of the beam. After passing through the polaroid, the beam remains polarized but in the plane of the second polaroid and its intensity is reduced, according to the Law of Malus, by the factor of 
This yields 
. Substituting from the previous step we get
yielding
and finally,
