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sweet [91]
2 years ago
8

An experiment is conducted in which red light is diffracted through a single slit. part a listed below are alterations made, one

at a time, to the original experiment, and the experiment is repeated. after each alteration, the experiment is returned to its original configuration. which of these alterations decreases the angles at which the diffraction minima appear? select all that apply. view available hint(s) select all that apply. a green, rather than red, light source is used. the slit width is halved. the slit width is doubled. the distance between the slits and the screen is doubled. the distance between the slits and the screen is halved. the experiment is conducted in a water-filled tank.
Physics
2 answers:
tensa zangetsu [6.8K]2 years ago
8 0

Color spectrum

<h2>Further Explanation </h2>

A spectrum is a condition or price that is not limited to a set price but can change indefinitely on a continuum. This word evolved from the Latin word, specter, which means ghosts, but the modern meaning now comes from its use in natural science. A spectrum is a condition or price that is not limited to a set price but can change indefinitely on a continuum. This word evolved from the Latin word, specter, which means ghosts, but the modern meaning now comes from its use in natural science. White is an impression that is captured by the eye and is a combination (superposition of several wavelengths) of various unique wavelengths.

For an easy example, I take the example of white can be produced from a combination of 3 wavelengths of Red Green and Blue or RGB (Red Green Blue).

Spectral is the result of the interaction between electromagnetic energy (EM) and an object. There are objects that have absorption properties (absorption) of high EM and low reflections, otherwise, there are objects that have low absorption and high reflectivity. These reflection and absorption patterns are different for different wavelengths. If it is associated with satellite images, then each object will provide a different EM reflection, so that we are able to distinguish an object from another object (identification). The instrument used to measure the reflectance value of an object is called a spectrometer. On the market there are spectrometers of higher spectral resolution, in addition, there are also spectrometers that can be used underwater (underwater).

the red color is produced from electromagnetic wavelengths/light in the range of 700-630 nanometers.

the green color is produced from electromagnetic/light wavelengths in the range of 560-490 nanometer meters.

the blue color is produced from electromagnetic/light wavelengths in the range of ~ 490-450 nanometer meters.

Learn more

Color spectrum brainly.com/question/12464893

Details

Grade:  High School

Subject:  Physics

keywords: Color spectrum

LenaWriter [7]2 years ago
6 0

Answer:

Where is the rest of the article to read form at?

Explanation:

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A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by
aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

x(t)=\alpha t^2 - \beta t^3

where

\alpha = 1.50 m/s^2

\beta = 0.05 m/s^3

The average velocity is given by the ratio between the displacement and the time taken:

v=\frac{\Delta x}{\Delta t}

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

The position at t = 2.00 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

So the displacement is

\Delta x = x(2)-x(0)=5.6-0=5.6 m

The time interval is

\Delta t = 2.0 s - 0 s = 2.0 s

And so, the average velocity in this interval is

v=\frac{5.6 m}{2.0 s}=2.8 m/s

b)

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

While the position at t = 4.00 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = x(4)-x(0)=20.8-0=20.8 m

The time interval is

\Delta t = 4.0 - 0 = 4.0 s

So the average velocity here is

v=\frac{20.8}{4.0}=5.2 m/s

c)

The position at t = 2 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

While the position at t = 4 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = 20.8 - 5.6 = 15.2 m

While the time interval is

\Delta t = 4.0 - 2.0 = 2.0 s

So the average velocity is

v=\frac{15.2}{2.0}=7.6 m/s

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

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6 0
2 years ago
Two resistors ( 3 ohms &amp; 6 ohms) in a series circuit with a power supply = 12 volts. The current through resistor 6 ohms is
Scorpion4ik [409]

In a series circuit . . .

-- The total resistance is the sum of the individual resistors.

-- The current is the same at every point in the circuit.

The total resistance in this circuit is (3Ω + 6Ω )  =  9Ω

The current at every point is (V/R) = (12v / 9Ω ) = <em>1.33 A</em> .

Pick choice<em> (a)</em>.

6 0
1 year ago
A car increases its speed from 9.6 meters per second to 11.2 meters per second in 4.0 seconds. The average acceleration of the c
Gwar [14]
The acceleration is the change of speed/velocity over time. Thus to calculate this you do (V1-V2)/T or (11.2-9.6)/4 or 0.4 m/s^2
5 0
2 years ago
Read 2 more answers
Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B
gregori [183]

Answer:

The force P required  is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

tan \ \alpha = \dfrac{CD}{BD}

where;

BD = AB-AD

Then;

tan \ \alpha = \dfrac{CD}{AB-AD}

\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )

replacing their respective values, where;

CD = 2 sin 30° m,  AB = 2m and AD = 2 cos 30° m

\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )

\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )

\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )

\alpha = tan^{-1} \bigg(3.73\bigg )

\alpha \simeq 75^0

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

T_{BC} = P

Using the force equilibrium expression along the horizontal direction.

\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0

replacing the value of \alpha \simeq 75^0

-T_{AC} \  cos 30^0 + P cos 75^0  = 0

P \ cos \ 75^0 = T_{AC} \ cos \ 30^0

P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - -  (1)

Along the vertical direction, the force equilibrium equation can be expressed as:

\sum F_y =0

-W + P \ sin \alpha + T_{AC} \ sin \ 30^0  = 0

W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0

replacing \alpha \simeq 75^0 and P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}

W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0

Also, replacing W for (200 × 9.81) N

200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0

200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0

1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)

1962= T_{AC} \ (0.8660\times 3.732 + 0.5)

1962= T_{AC} \ (3.231912 + 0.5)

1962= T_{AC} \ (3.731912)

T_{AC}  = \dfrac{1962}{ \ (3.731912)}

T_{AC}  = 525.736 \ N

From P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}

P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}

P =\dfrac{ 525.736 \times0.866}{0.2588}

P = 1759.22 N

Thus, the force P required  is 1759.22 N

6 0
1 year ago
A professor designing a class demonstration connects a parallel-plate capacitor to a battery, so that the potential difference b
Lesechka [4]

Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

        C = Q / V

        Q = C V

         

can also be calculated using geometry consideration

        C = e or A / d

         

we reduce to the SI system

       A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

       d = 1.53 cm = 1.53 10⁻² m

we substitute

         Q = eo A / d V

         Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

         Q = 3.9757 10⁻¹⁰ C

         

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

           V = Vo / k

            V = 275/80

            V = 3.4375 V

c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

             U = 5.47 10⁻⁸ J

let's reduce to nJ

              109 nJ = 1 J

               U = 54.7 nJ

d) energy after submerging

             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

             U = U₀ / k

             U = 54.7 / 80 nJ

              U = 0.68375 nJ

the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

           

6 0
2 years ago
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