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mario62 [17]
3 years ago
6

In a Young's double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separatio

n between adjacent bright fringes on a screen 5 m from the slits is: CONVERT FIRST
Physics
1 answer:
asambeis [7]3 years ago
4 0

Answer:

Δx = 2.5 x 10⁻³ m = 2.5 mm

Explanation:

The distance between two consecutive fringes, also known as fringe spacing, in Young's Double Slit Experiment, is given as follows:

Δx = λL/d

where,

Δx = distance between consecutive fringes = ?

λ = wavelength of light = 500 nm = 5 x 10⁻⁷ m

L = Distance between slits and screen = 5 m

d = slit separation = 1 mm = 1 x 10⁻³ m

Therefore,

Δx = (5 x 10⁻⁷ m)(5 m)/(1 x 10⁻³ m)

<u>Δx = 2.5 x 10⁻³ m = 2.5 mm</u>

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C. Echolocation

Echolocating animals emit calls out to the environment and listen to the echoes of those calls that return from various objects near them. They use these echoes to locate and identify the objects.
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A spring stretches by 0.0177 m when a 2.82-kg object is suspended from its end. How much mass should be attached to this spring
Artemon [7]

The mass attached to the spring must be 0.72 kg

Explanation:

The frequency of vibration of a spring-mass system is given by:

f=\frac{1}{2\pi} \sqrt{\frac{k}{m}} (1)

where

k is the spring constant

m is the mass attached to the spring

We can find the spring constant by using Hookes' law:

F=kx

where

F is the force applied on the spring

x is the stretching of the spring

When a mass of m = 2.82 kg is applied to the spring, the force applied is the weight of the mass, so we have

mg=kx

and using g=9.8 m/s^2 and x=0.0177 m, we find

k=\frac{mg}{x}=\frac{(2.82)(9.8)}{0.0177}=1561.3 N/m

Now we want the frequency of vibration to be

f = 7.42 Hz

So we can rearrange eq.(1) to find the mass m that we need to attach to the spring:

m=\frac{k}{(2\pi f)^2}=\frac{1561.3}{(2\pi (7.42))^2}=0.72 kg

#LearnwithBrainly

6 0
3 years ago
The diagram shows a state of matter in a closed system before and after undergoing a
belka [17]

Answer:

the second one

Explanation:

3 0
3 years ago
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You hear three beats per second when two sounds tones are generated. The frequency of one tone is known to be 610 Hz. The freque
tiny-mole [99]

The frequency of the other wave is 613 Hz or 607 Hz.

The difference between the frequencies of two waves is called the beat frequency.

Here, one wave has a frequency 610 Hz and the beat frequency is 3 beats per second.

Which has a higher frequency is not mentioned. Therefore, there are two possibilities.

Δf = | 610 - 613 | = 3

or

Δf = | 610 - 607 | = 3

Therefore, the frequency of the other wave is 613 Hz or 607 Hz.

Learn more about beat frequency here:

brainly.com/question/14157895

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2 years ago
Find the equilibrant of two 10.0-N forces acting upon a body when the angle between the forces is 90° Solve graphically using a
bazaltina [42]

The equilibrant force of the two given forces is 14.14 N.

<h3 /><h3 /><h3>What is equilibrant force?</h3>
  • This is a single force that balances other given forces.

The given parameters:

  • First force, F₁ = 10 N
  • Second force, F₂ = 10 N
  • Angle between the forces, θ = 90⁰

The equilibrant force of the two given forces is calculated as follows;

F = \sqrt{F_1 ^2 + F_2 ^2} \\\\F = \sqrt{(10)^2 + (10)^2} \\\\F = 14.14 \ N

Thus, the equilibrant force of the two given forces is 14.14 N.

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