Answer:
0.5
Explanation:
Data provided in the question:
The angle between their transmission axes, θ = 60°
Now,
We have the relation,
I₁ = I₀cos²θ
where,
I₁ is the intensity of the transmitted light
I₀ is the intensity of the incident light
on rearranging, we get
=cos²60°
or
=0.5
If the field is in a vacuum, the magnetic field is the dominant factor determining the motion. Since the magnetic force is perpendicular to the direction of travel, a charged particle follows a curved path in a magnetic field. The particle continues to follow this curved path until it forms a complete circle. Another way to look at this is that the magnetic force is always perpendicular to velocity, so that it does no work on the charged particle. The particle’s kinetic energy and speed thus remain constant. The direction of motion is affected but not the speed.
A negatively charged particle moves in the plane of the paper in a region where the magnetic field is perpendicular to the paper (represented by the small × ’s—like the tails of arrows). The magnetic force is perpendicular to the velocity, so velocity changes in direction but not magnitude. The result is uniform circular motion.
Answer:
A measured force of (46.5 0.8 N ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N )
Explanation:
From the question we are told that
Measured force is ![F_M = [46.5 \pm 0.8 \ N ]](https://tex.z-dn.net/?f=F_M%20%20%3D%20%20%5B46.5%20%5Cpm%200.8%20%5C%20%20N%20%5D)
Calculated force is ![F_c = [48.4 \pm 0.6 \ N ]](https://tex.z-dn.net/?f=F_c%20%3D%20%20%5B48.4%20%5Cpm%200.6%20%5C%20%20N%20%5D)
Generally the measured force in interval form is

=> 
Generally the calculated force in interval form is

=> 
Generally looking both interval we see that they do not intersect at any point Hence
A measured force of (46.5 0.8 N ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N )