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Juli2301 [7.4K]
3 years ago
11

it's nighttime, and you've dropped your goggles into a swimming pool that is 3.2 m deep. If you hold a laser pointer 1.0 m above

the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.5 m from the edge. How far are the goggles from the edge of the pool?

Physics
1 answer:
Scrat [10]3 years ago
3 0

Answer:

5.2 m

Explanation:

from the question we are given the following

depth of pool (d) = 3.2 m

height of laser above the pool (h) = 1 m

point of entry of laser beam from edge of water (l) = 2.5 m

we first have to calculate the angle at which the laser beam enters the water (∝),

tan ∝ = \frac{1}{2.2}

∝ = 24.44 degrees

from the attached diagram, the angle with the normal (i) = 90 - 24.4 = 65.56 degrees

lets assume it is a red laser which has a refractive index of 1.331 in water, and with this we can find the angle of refraction (r) using the formula below

refractive index = \frac{sin i}{sin r}

1.331 = \frac{sin 65.56}{sin r}

r = 43.16 degrees

we can get the distance (x) from tan r = \frac{x}{3.2}

tan 43.16 = \frac{x}{3.2}

x = 3 m

To get the total distance we need to add the value of x to 2.2 m

total distance = 3 + 2.2 = 5.2 m

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A horizontal vinyl record of mass 0.10 kg and radius 0.10 m rotates freely about a vertical axis through its center with an angu
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The angular speed of the record is 3.36 rad/s.

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Given that,

Mass of record= 0.10 kg

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Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree
kvv77 [185]

Answer:

1

When second polarizer is removed the intensity after it passes through the stack is    

                    I_f_3 = 27.57 W/cm^2

2 When third  polarizer is removed the intensity after it passes through the stack is    

                I_f_2 = 102.24 W/cm^2

Explanation:

  From the question we are told that

       The angle of the second polarizing to the first is  \theta_2 = 21^o  

        The angle of the third  polarizing to the first is     \theta_3 = 61^o

        The unpolarized light after it pass through the polarizing stack   I_u = 60 W/cm^2

Let the initial intensity of the beam of light before polarization be I_p

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

                     I_1 = \frac{I_p}{2}

Now according to Malus’ law the  intensity of light that would emerge from the second polarizing filter is mathematically represented as

                    I_2 = I_1 cos^2 \theta_1

                       = \frac{I_p}{2} cos ^2 \theta_1

The intensity of light that will emerge from the third filter is mathematically represented as

                  I_3 = I_2 cos^2(\theta_2 - \theta_1 )

                          I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]

making I_p the subject of the formula

                  I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}

    Note that I_u = I_3 as I_3 is the last emerging intensity of light after it has pass through the polarizing stack

         Substituting values

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}

                      I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}

                           =234.622W/cm^2

When the second    is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

                      I_f_3 = \frac{I_p}{2} cos ^2 \theta_2

I_f_3 is the intensity of the light emerging from the stack

                     

substituting values

                     I_f_3 = \frac{234.622}{2} * cos^2(61)

                       I_f_3 = 27.57 W/cm^2

  When the third polarizer is removed  the  second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be  

                  I_f_2 = \frac{I_p}{2} cos ^2 \theta_1

I_f_2 is the intensity of the light emerging from the stack

Substituting values

                  I_f_2 =  \frac{234.622}{2} cos^2 (21)

                     I_f_2 = 102.24 W/cm^2

   

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The total resistance is R = voltage / current

This resistance R = (L/A)r where:
r is the resistivity
L is the length of the conductor
A is the cross-sectional area of the conductor.

thus r = RA/L

You are given L; you can compute R from the voltage and current you are given; and the cross-sectional area of a round wire is (pi)(radius^2) or (pi/4)(diameter^2)
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