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GREYUIT [131]
2 years ago
13

At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to

Physics
1 answer:
mote1985 [20]2 years ago
6 0

Answer:

Ok, the question is incomplete buy ill try to answer this in a general way.

Suppose that you have no-polarized light.

When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.

Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°

and:

I(0°) = I0*cos^2(0°) = I0

So the intensity does not change.

Now, knowing the initial intensity, you can find the angle needed to get a given intensity.

For example, if the question was:

"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"

We should solve:

I(θ) = A = I0*cos^2(θ)

(A/i0) = cos^2(θ)

√(A/I0) = cos(θ)

Acos(√(A/I0)) = θ

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The electric potential in a particular region of space varies only as a function of y-position and is given by the function V(y)
nikdorinn [45]

Answer:

E = 55.9583\ Volts/meter

Explanation:

First let's find the electric potential using y = 22.5:

V(y) = 1.69y^2 +15.6y+52.5

V(22.5) = 1.69(22.5)^2 + 15.6*22.5 + 52.5

V(22.5) = 1259.0625\ Volts

Then, to find the magnitude of the electric field, we just need to divide the electric potential by the distance y:

E = V/d

E = 1259.0625/22.5

E = 55.9583\ Volts/meter

3 0
2 years ago
How, if at all, would the equations written in Parts C and E change if the projectile was thrown from the cliff at an angle abov
sveta [45]

Answer:

x = v₀ cos θ   t ,   y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This is a projectile launch exercise, in this case we will write the equations for the x and y axes

Let's use trigonometry to find the components of the initial velocity

              sin θ = v_{oy} / v₀

              cos θ = v₀ₓ / v₀

              v_{y} = v_{oy} sin θ

              v₀ₓ = vo cos θ

now let's write the equations of motion

X axis

         x = v₀ₓ t

         x = v₀ cos θ   t

        vₓ = v₀ cos θ

Y axis

        y = y₀ + v_{oy} t - ½ g t2

        y = y₀ + v₀ sin θ t - ½ g t2

        v_{y} = v₀ - g t

       v_{y} = v₀  sin θ - gt

        v_{y}^{2} = v_{oy}^2 sin² θ - 2 g y

As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components

7 0
2 years ago
An electric eel (Electrophorus electricus) can produce a shock of up to 600 V and a current of 1 A for a duration of 2 ms, which
Irina-Kira [14]

Answer:

2\times 10^{-3}\ C

6000

1.2 J

3.33\times 10^{-6}\ F

Explanation:

I = Current = 1 A

t = Time = 2 ms

n = Number of electrocyte

V = Voltage = 100 mV

Charge is given by

Q=It\\\Rightarrow Q=1\times 2\times 10^{-3}\\\Rightarrow Q=2\times 10^{-3}\ C

The charge flowing through the electrocytes in that amount of time is 2\times 10^{-3}\ C

The maximum potential is given by

V_m=nV\\\Rightarrow n=\dfrac{V_m}{V}\\\Rightarrow n=\dfrac{600}{100\times 10^{-3}}\\\Rightarrow n=6000

The number of electrolytes is 6000

Energy is given by

E=Pt\\\Rightarrow E=V_mIt\\\Rightarrow E=600\times 1\times 2\times 10^{-3}\\\Rightarrow E=1.2\ J

The energy released when the electric eel delivers a shock is 1.2 J

Equivalent capacitance is given by

C_e=\dfrac{Q}{V_m}\\\Rightarrow C_e=\dfrac{2\times 10^{-3}}{600}\\\Rightarrow C_e=3.33\times 10^{-6}\ F

The equivalent capacitance of all the electrocyte cells in the electric eel is 3.33\times 10^{-6}\ F

8 0
2 years ago
If you drive through water, your brakes may become slippery and ineffective. To dry the brakes off, __________.
GenaCL600 [577]

Answer:

I am not a driver, but I think it's C.

Explanation:

6 0
2 years ago
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qwelly [4]

14250. I just took it

4 0
2 years ago
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