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34kurt
3 years ago
10

The angle between the axes of two polarizing filters is 45.0^\circ45.0 ​∘ ​​ . By how much does the second filter reduce the int

ensity of the light coming through the first? Select the correct answer 0.750 0.800 0.250 0.500 Your Answer 0.125 Saved 1 of 3 attempts used Part b (1 points) What now is the direction of polarization of the transmitted light? (You can assume the first polarizer defines the angle \theta = 0θ=0.) Select the correct answer
Physics
1 answer:
suter [353]3 years ago
7 0

Answer

given,                                                                      

angle between two polarizing filters = 45°

filter reduce intensity = ?                          

a) I = I₀ Cos² θ                                

here θ = 45⁰                                

I = \dfrac{I_0}{2}                      

intensity of the light is reduced by 0.500

correct answer from the given option D

b) direction of the polarization                    

                        θ = 45°                  

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Alekssandra [29.7K]

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

F = BIl \ sin(\theta)

(a) When the angle, θ = 60 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N

(b) When the angle, θ = 90 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N

(c) When the angle, θ = 120 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N

6 0
2 years ago
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Answer:

The half-life is t_{1/2} = 1.005 h

Explanation:

Using the decay equation we have:

A=A_{0}e^{-\lambda t}

Where:

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We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that A = \frac{A_{0}}{2}

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Taking the natural logarithm on each side we have:

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Now, the relationship between the decay constant λ and the half-life t(1/2) is:

\lambda = \frac{ln(2)}{t_{1/2}}

t_{1/2} = \frac{ln(2)}{\lambda}

t_{1/2} = \frac{ln(2)}{0.69}

t_{1/2} = 1.005 h

I hope it helps you!

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