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KengaRu [80]
3 years ago
12

PLEASE I NEED THUS THX SOOOOOO MUCH IT MEANS A LOT! THIS WILL KILL MY GRADE IF ITS BAD Properties of Light Lab Report Instructio

ns: In the Properties of Light—Bending Light Lab you will explore how light rays interact as they pass from one material, or medium, to another. Record your observations in the lab report below. You will submit your completed report. Name and Title: Include your name, instructor's name, date, and name of lab. Objective(s): In your own words, what was the purpose of this lab? Hypothesis: In this section, please include the if/then statement you developed during your lab activity. This statement reflects your predicted outcome for the experiment. If I project light rays through (choose one: air, water, glass), then they will refract, or bend, the most. Procedure: Please be sure to identify the test variable (independent variable) and the outcome variable (dependent variable) for this investigation. Remember, the test variable is what is changing in this investigation. The outcome variable is what you are measuring in this investigation. Test variable (independent variable): Outcome variable (dependent variable): Select Intro to begin. Once the simulation loads, you will drag the protractor onto the screen so that the 0 lines up with the dotted line in the center of the screen. You will test all three materials in the bottom, blue half of the screen: air, water, and glass. Record your observations of how the light rays interact as they pass from air, on the top of the screen, through the three materials on the bottom of the screen Data: Record the data from each trial in the data chart below. Be sure to fill in the chart completely. Material Interactions List any observations you made as the light rays passed from the air on the top of the screen to the selected material on the bottom of the screen. Evidence Using the protractor, measure the angle as the light passes from the top material to the bottom material on the screen. Also, note the index of refraction for each as indicated on the screen. Trial One Air on top Angle: Index of refraction: Air on bottom thxxxxxx do much u made my day​
Physics
1 answer:
adell [148]3 years ago
6 0

Answer:

Objective(s):

In your own words, what was the purpose of this lab?

To explain how different materials make light bend.  

Hypothesis:

In this section, please include the if/then statement you developed during your lab activity. This statement reflects your predicted outcome for the experiment.

If I project light rays through (choose one: air, water, glass), then they will refract, or bend, the most.

If I project light rays through glass, then they will refract the most.  

Procedure:

Please be sure to identify the test variable (independent variable) and the outcome variable (dependent variable) for this investigation. Remember, the test variable is what is changing in this investigation. The outcome variable is what you are measuring in this investigation.

Test variable (independent variable):

Outcome variable (dependent variable):

Select Intro to begin.

Once the simulation loads, you will drag the protractor onto the screen so that the 0 lines up with the dotted line in the center of the screen.

You will test all three materials in the bottom, blue half of the screen: air, water, and glass.

Record your observations of how the light rays interact as they pass from air, on the top of the screen, through the three materials on the bottom of the screen

Data:

Record the data from each trial in the data chart below. Be sure to fill in the chart completely.

Material

Interactions

List any observations you made as the light rays passed from the air on the top of the screen to the selected material on the bottom of the screen.

Evidence

Using the protractor, measure the angle as the light passes from the top material to the bottom material on the screen. Also, note the index of refraction for each as indicated on the screen.

Trial One

Air on top

The light goes straight through the air  

Angle:  

Acute  

Index of refraction:

1.00

Air on bottom

 

Trial Two

Air on top

The light bends in the water  

Angle:

Actute  

Index of refraction:

1.34  

Water on bottom

 

Trial Three

Air on top

The light almost goes straight down when it hits the glass  

Angle:

Acute  

Index of refraction:

1.60  

Glass on bottom

 

Conclusion:

Your conclusion will include a summary of the lab results and an interpretation of the results. Please write in complete sentences.

Did your data support your hypothesis? Use evidence to explain.

Yes it did, the glass index refraction (1.60) is higher than the other ones.

Which material refracted the light rays the most: air, water, or glass?

glass

Which material refracted the light rays the least: air, water, or glass?

air

How does density affect refraction?

It reflects light.

Diamonds are a very dense material. Predict what would happen to the light ray if you projected it from air through a diamond.

The light would reflect everywhere,

Explain where you observe reflection, refraction, and absorption of light in your everyday activities.

In mirrors and when light from a window hits a glass of water.  

Explanation:

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Answer:

c. Joints allow the roadway to expand and contract as cars put force on the bridge

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g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p
kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r = \frac{mv}{qB}

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

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We can use the work-energy theorem to solve.

Recall that:

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The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

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Now, plug in the values:

786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta

Solve for theta:

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