Answer:
θ_p = 53.0º
Explanation:
For reflection polarization occurs when a beam is reflected at the interface between two means, the polarization in total when the angle between the reflected and the transmitted beam is 90º
Let's write the transmission equation
n1 sin θ₁ = ne sin θ₂
The angle to normal (vertcal) is
180 = θ2 + 90 + θ_p
θ₂ = 90 - θ_p
Where θ₂ is the angle of the transmitted ray θ_p is the angle of the reflected polarized ray
We replace
n1 sin θ_p = n2 sin (90 - θ_p)
Let's use the trigonometry relationship
Sin (90- θ_p) = sin 90 cos θ_p - cos 90 sin θ_p = cos θ_p
In the law of reflection incident angle equals reflected angle,
ni sin θ_p = ns cos θ_p
n₂ / n₁ = sin θ_p / cos θ_p
n₂ / n₁ = tan θ_p
θ_p = tan⁻¹ (n₂ / n₁)
Now we can calculate it
The refractive index of air is 1 (n1 = 1) the refractive index of seawater varies between 1.33 and 1.40 depending on the amount of salts dissolved in the water
n₂ = 1.33
θ_p = tan⁻¹ (1.33 / 1)
θ_p = 53.0º
n₂ = 1.40
θ_p = tan⁻¹ (1.40 / 1)
Tep = 54.5º
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Answer:
R = 7 [amp]
Explanation:
To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance. In this way, we have the following equation.
V = I*R
where:
V = voltage = 49 [V] (units of volts)
I = current = 7 [amp] (amperes)
R = resistance [ohms]
Now clearing R.
R =V/I
R = 49/7
R = 7 [amp]
Answer:
bubble wrap in stuff animal
Explanation:
did it
Answer:
Yes, the fuse will blow.
Explanation:
We'll begin by converting 50000mA to amperes (A) .this is illustrated below:
1000mA = 1A
Therefore, 50000mA = 50000/1000 = 50A.
From the question given above, the plug can only accommodate 5A. Now if 50000mA i.e 50A is passed through the plug, the plug will blow because the 50A is higher than what the plug can accommodate.
