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motikmotik
1 year ago
10

If the incident intensity of the light is 26 W/m, what is the intensity of the light that emerges from the filter

Physics
1 answer:
REY [17]1 year ago
4 0

47W/m^{2}  is the intensity of the light that emerges from the filter

Use Malus's law, the intensity of the light is,

        I=I_{0}, cos² ∅

The intensity of the beam from the first polarizer is equal to the half of

the initial intensity.

            I_{1} = I_{0}/2

Substitute the numerical values we get

94 W/m² 2

= 47 W/m²

What is intensity ?

In physics, the power transferred per unit area is known as the intensity or flux of radiant energy, where the area is measured on a plane perpendicular to the direction of the energy's propagation. Watts per square meter (W/m2) and kilograms per square meter (kg/s3) are the units used in the SI system. With waves like acoustic waves (sound) or electromagnetic waves like light or radio waves, intensity is most usually employed to describe the average power transfer across one period of the wave. Other situations where energy is exchanged can also be described in terms of intensity. One could, for instance, figure out how much kinetic energy each drop of water from a sprinkler is carrying.

To learn more about intensity visit:

brainly.com/question/25556938

#SPJ4

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A person runs 15 kilometers in two hours what is his/her average speed ​
bazaltina [42]

Answer:

\displaystyle V=7.5\ km/h

Explanation:

<u>Average Speed </u>

If an object travels a distance d in a time t regardless of the direction, the average speed is the quotient of the distance over the time:

\displaystyle V=\frac{d}{t}

It's known a person runs d=15 kilometers in t=2 hours, thus his/her average speed is:

\displaystyle V=\frac{15\ km}{2\ h}

Calculating:

\boxed{\displaystyle V=7.5\ km/h}

5 0
3 years ago
What is a rarefaction?
shutvik [7]

Answer:

Rarefaction is the reduction of an item's density, the opposite of compression. Like compression, which can travel in waves, rarefaction waves also exist in nature. A common rarefaction wave is the area of low relative pressure following a shock wave.

Explanation:

8 0
3 years ago
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A person standing on a scale feels a normal force of 655 N pushing up on him. What is his mass? (Unit=kg)
dimulka [17.4K]

Answer:

Approximately 66,8kg

Hope this help :D

Explanation:

F=mg. So 655N= 9.8 m/s² × mass.

Mass = 655 / 9.8

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3 years ago
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A 0.9 kg ball attached to a cord is whirled in a vertical circle of radius 2.5 m. Find the minimum speed needed at the top of th
lapo4ka [179]

Answer:

The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.

Explanation:

By the Principle of Energy Conservation we understand that the minimum speed needed by the ball is that speed such that maximum height reached is equal to the diameter of the vertical circle, that is:

K =U_{g} (1)

Where:

K - Translational kinetic energy, measured in joules.

U_{g} - Gravitational potential energy, measured in joules.

By definitions of translational kinetic and gravitational potential energies, we expand the equation above and clear the initial speed of the ball:

\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g\cdot h

v = \sqrt{2\cdot g\cdot h} (2)

Where:

m - Mass, measured in kilograms.

v - Initial speed, measured in meters per second.

g - Gravitational acceleration, measured in meters per square second.

h - Maximum height of the ball, measured in meters.

If we know that g = 9.807\,\frac{m}{s^{2}} and h = 5\,m, then the initial speed of the ball is:

v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}

v\approx 9.903\,\frac{m}{s}

The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.

3 0
3 years ago
I need help with one through six please
dybincka [34]

Answer:

1] 8500000 = <u>8.5 × 10⁶</u>

2] .000072 = <u>7.2 × 10⁻⁵</u>

3] 5.3 × 10⁴ = <u>53000</u>

4] 2.8 × 10⁻³ = <u>0.0028</u>

5] Velocity = \frac{distance}{time}

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6] Acceleration = \frac{V1-V2}{time}

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 A = \frac{15}{3}

 <u>A = 3 m/s²</u>

7 0
2 years ago
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