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elena55 [62]
1 year ago
9

monochromatic light from a distant source is incident on a slit 0.75 mm wide. on screen 2 m away, the distance from the central

maximum of the diffraction pattern to the first minimum is measured to be 1.35 mm. find the wavelength of the light.
Physics
1 answer:
hjlf1 year ago
5 0

Displacement from the center line for minimum intensity is 1.35 mm , width of the slit  is 0.75 so  Wavelength of the light  is 506.25.

<h3>How to find Wavelength of the light?</h3>

When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.

This is a straightforward situation in which we can apply the

Fraunhofer single slit diffraction equation:

y = mλD/a

Where:

y = Displacement from the center line for minimum intensity =  1.35 mm

λ =  wavelength of the light.

D = distance

a = width of the slit = 0.75

m = order number = 1

Solving for λ

λ = y + a/ mD

Changing the information that the issue has provided:

λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2  

=5.0625 *10^-7 = 506.25

so

Wavelength of the light 506.25.

To learn more about Wavelength of the light refer to:

brainly.com/question/15413360

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Resolving vectors question. See diagram enclosed.

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3 years ago
Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0 °C for the following reacti
Yakvenalex [24]

The equilibrium constant of the reaction at 25⁰c will be 426827.5.

<u />

  • <u>Theory-</u>

<u>Equilibrium constant</u> :The equilibrium constant comes from the chemical equilibrium law. For the chemical equilibrium state, at a fixed constant temperature, the ratio of the product of the reaction's multiplication to the concentration of its reactants' multiplication, and each is raised to the power to the corresponding coefficients of the elements in the reaction.

The chemical equilibrium is given by for a general chemical reaction.

a. A+ b. B ⇌ c. C+ d. D,.

Kc =[C]c [D]d/[A]a [B]b.

<u>Gibb's free energy</u> :The second law of thermodynamics can be arranged in such a way that it gives a new expression when a chemical reaction happens at a constant temperature and constant pressure.

G=H-TS

  • <u>Calculations</u>:-

T=25⁰c

G=51.4 x 10³J

\\\\k=GR+\frac{nRT}{Z} \\

k= equilibrium constant ,G=Gibbs free energy ,n= no. of moles ,R=Gas constant ,T=temperature ,Z=compressibility

Ideal.Situation=\left \{ {{Z=1} \atop {n=1}} \right.

\\\\\\k=GR+RT

k=51.4 x 10³ x 8.3 + 8.3 x 25

k=426827.5

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6 0
1 year ago
2. While standing near a bus stop, a student hears a distant horn beeping. The frequency emitted by the horn is 440 Hz. The bus
jarptica [38.1K]

Given Information:

Frequency of horn = f₀ = 440 Hz

Speed of sound = v = 330 m/s

Speed of bus = v₀ = 20 m/s

Answer:

Case 1. When the bus is crossing the student = 440 Hz

Case 2. When the bus is approaching the student = 414.9 Hz

Case 3. When the bus is moving away from the student = 468.4 Hz

Explanation:

There are 3 cases in this scenario:

Case 1. When the bus is crossing the student

Case 2. When the bus is approaching the student

Case 3. When the bus is moving away from the student

Let us explore each case:

Case 1. When the bus is crossing the student:

Student will hear the same frequency emitted by the horn that is 440 Hz.

f = 440 Hz

Case 2. When the bus is approaching the student

f = f₀ ( v / v+v₀ )

f = 440 ( 330/ 330+20 )

f = 440 ( 330/ 350 )

f = 440 ( 0.943 )

f = 414.9 Hz

Case 3. When the bus is moving away from the student

f = f₀ ( v / v+v₀ )

f = 440 ( 330/ 330-20 )

f = 440 ( 330/ 310 )

f = 440 ( 1.0645 )

f = 468.4 Hz

6 0
3 years ago
On a hot day, the deck of a small ship reaches a temperature of 48
AlekseyPX

The final temperature of the seawater-deck system is 990°C.

<h3>What is heat?</h3>

The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.

The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.

Suppose for 1 kg of sea water, the heat transferred from the system is given by

3,710,000 = 1 x 3,930 x (T - 48.17)

T = 990°C  to the nearest tenth.

The final temperature of the seawater-deck system is 990°C.

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