The average depth of water in a place near the coast of the Pacific ocean is 5.0 The largest tidal range in that place is 2.0m C
1 answer:
A sinusoidal function is a function based on the sine function.
The maximum depth of the water at the place, is <u>6.0 meters</u>.
Reason:
The known parameters are;
The average depth of water at a place, = 5.0
The largest tidal range in the place, R = 2.0 m.
Required;
To calculate the maximum depth of water at the given place.
Solution:
The tidal range is the consecutive difference between highs and lows of
the tides, which is a sinusoidal motion.
Therefore, the tidal range is twice the amplitude of the tide level.
The amplitude of the tide, <em>a</em> = m = 1 m
The maximum depth of water at the place, = a +
∴ = 1.0 + 5.0 = 6.0
The maximum depth, = <u>6.0 meters</u>
Learn more about sinusoidal functions here:
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Answer:
a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º
Explanation:
a) For this exercise we must use Malus's law
I = I₀ cos² θ
where tea is the angle between the two polarizers.
We apply this expression to our case
* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical
I₁ = I₀ cos² θ
* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.
The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical
θ₂ = 90- θ’ = θ
fiate that in the exercise we must take a reference system and measure everything with respect to this system.
I = I₁ cos² θ'
we substitute
I = (I₀ cos² tea) cos² (θ - 90)
cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ
I = Io cos² θ sin² θ
1= cos²θ+ sin²θ
sin²θ = 1 - cos²θ
I= I₀ (cos²θ - cos⁴θ)
b) to find when the intensity is maximum,
we can use that we have an extreme point when the drift is zero
= 0
\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0
whereby
cos θ - 2 cos³ θ = 0
cos θ ( 1 - 2 cos² θ) = 0
The zeros of this function are in
θ = 90º
1-2cos²θ =0 cos θ = 0.25 θ = 75.5º
Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.
Consequently, the angle that allows the maximum intensity to pass is 75.5º