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Andrews [41]
3 years ago
10

Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po

larizing filter whose axis is at 45.0 ∘∘ to that of the first. Part A Determine the intensity of the beam after it has passed through the second polarizer. Enter the factor only. For example, enter 0.2500.250, that means your answer is I=0.250I0I=0.250I0. II = nothing I0I0 SubmitRequest Answer Part B Determine its state of polarization. Determine its state of polarization. The light is linearly polarized along the axis of the first filter. The light is linearly polarized along the axis of the second filter. The light is linearly polarized perpendicular to the axis of the second filter. The light is linearly polarized perpendicular to the axis of the first filter. SubmitRequest Answer Provide Feedback
Physics
1 answer:
Montano1993 [528]3 years ago
3 0

Answer:

A) I = Io 0.578,   B) he light that leaves the polarized is completely polarized, being perpendicular to the axis of the second filter

Explanation:

A) Light passing through a polarizer must comply with the / bad law

          I = Io cos2 tea

Where is at the angle of the polarizer and incident light

          I = Io cos2 45

         I = Io 0.578

Therefore the beam intensity is 0.578 of the incident intensity

.B) the light that leaves the polarized is completely polarized, being perpendicular to the axis of the second filter

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Ultrasonic images are obtained from the inside organs of our body. This process uses which property of sound wave?
Temka [501]

This question involves the concepts of echo, ultrasonic images, ultrasonic sound waves.

The process of ultrasonic images uses the "echo" property of the sound waves.

Echo is the property of the sound wave by the virtue of which the sound wave reflects back to the source of the sound after hitting a surface or an object.

Ultrasonic images are obtained from inside organs of our body. This process involves the use of ultrasonic sound waves that have a frequency greater than 20,000 Hz. These sound waves are out of the range of audible sound by the human ear. When these ultrasonic sound waves are sent in form of pulses into the human body by the use of probes, they reflect back from the tissues of different organs to the probe. The probe then records the reflection properties of these sound waves and displays them in form of an image, known as ultrasonic images.

Learn more about echo here:

brainly.com/question/14335186?referrer=searchResults

The attached picture shows the process of ultrasonic imaging.

4 0
2 years ago
A force of 3kN acts on a car to make it accelerate by 1.5m/s/s. What is the mass of the car?
Masja [62]

Answer:

2

Explanation:

To find force it's force = mass times acceleration so to find mass you would divide force by acceleration

5 0
3 years ago
You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experimen
Nikolay [14]

Answer:

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center

Explanation:

Let  n₁ and n₂ be no of lines per unit length  of grating A and B respectively.

λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,

Distance of first maxima for green light

= λ₁ D/ d₁

Distance of first maxima for red light

= λ₂ D/ d₂

Given that

λ₁ D/ d₁ = λ₂ D/ d₂

λ₁ / d₁ = λ₂ / d₂

λ₁ / λ₂  = d₁ / d₂

But

λ₁  <  λ₂

d₁ < d₂

Therefore no of lines per unit length of grating A will be more because

no of lines per unit length  ∝ 1 / d

If grating B is illuminated with green light first maxima will be at distance

λ₁ D/ d₂

As λ₁ < λ₂

λ₁ D/ d₂ < λ₂ D/ d₂

λ₁ D/ d₂ < 1 m

In this case position of first maxima will be less than 1 meter.

Option a is correct .

5 0
3 years ago
A sand mover at a quarry lifts 2,000 kg of sand per minute a vertical distance of 12 m. The sand is initially at rest and is dis
Marianna [84]

Answer:

<h2>E. 3.95kW</h2>

Explanation:

Power is defined as the rate of workdone.

Power = Workdone/time taken

Given Workdone = Force * distance

Power = Force * distance/time taken

Power = mgd/t (F = mg)

m = mass of the sand in kg

g = acceleration due to gravity in m/s²

d = vertical distance covered in metres

t = time taken in seconds

Given m = 2000kg, d = 12m, t = 1min = 60secs, g = 9.8m/s²

Power = 2000*9.8*12/60

Power = 3920Watts

Minimum rate of power that must be supplied to this machine is 3920Watts or 3.92kW

5 0
3 years ago
When y⁷ is multiplied by y⁹ the answer is?
Allushta [10]

Answer:

y^16

Explanation:

who need to add the exponents only

7 + 9 = 16

therefore, the answer is y^16

5 0
2 years ago
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