The answer is letter C. 1.2 10^-11 N up
Solution:
F= Bqvsin(theta)
theta = sin 90 = 1
F= 1.4 T * 1.6x10^-19 * 5.2x10^7 ms^-1
F= 1.16 x 10^-11 N
Then the direction is upward.
Answer:
a) The angle of refraction is approximately 34.7
b) The angle the light have to be incident to give an angle of refraction of 90° is approximately 53.42°
Explanation:
According to Snell's law, we have;

The refractive index of the glass, n₁ = 1.66
The angle of incident of the light as it moves into water, θ₁ = 27.2°
a) The refractive index of water, n₂ = 1.333
Let θ₂ represent the angle of refraction of the light in water
By plugging in the values of the variables in Snell's Law equation gives;


θ₂ = arcsin(0.5692292265) ≈ 34.7°
The angle of refraction of the light in water, θ₂ ≈ 34.7°
b) When the angle of refraction, θ₂ = 90°, we have;


θ₁ ≈ arcsin(0.803) ≈ 53.42°
The angle of incident, θ₁, that would give an angle of refraction of 90° is θ₁ ≈ 53.42°
True, I'm not the best when it comes to science, but I'm pretty sure it's this
Answer:
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