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suter [353]
2 years ago
13

A laboratory experiment produces a double-slit interference pattern on a screen. If the screen is moved farther away from the sl

its, the fringes will be.
Physics
1 answer:
Mariana [72]2 years ago
7 0

The fringes will be farther apart

The width of the centre maxima on a screen is twice the distance between the dark fringes on either side at a distance L away: The pattern of light spreads out as you get further away from the screen (bigger L). The pattern of light widens as the hole gets smaller (smaller a).

<h3>What is Double slit experiment ?</h3>

The double-slit experiment involves directing a beam of light at a wall that has two vertical slits. The pattern that results from the light passing through the slits is captured on a photographic plate. A single line of light is seen when one slit is covered, aligned with the open slit.

  • In essence, waves passing through two closely spaced, parallel slits produce an interference pattern on a screen. Everyone should know this. This is true for all types of waves sound wave, light wave etc

Learn more about Double slit experiment here:

brainly.com/question/24196709

#SPJ4

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The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha
saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm =  4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0
3 years ago
. Which of the following is NOT a property of water? A. Ability to dissolve many substances B. Constant volume upon freezing C.
ss7ja [257]

Answer:

a

Explanation:

water is only able to desolve two substances , salt , sugar,

7 0
3 years ago
Read 2 more answers
According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f
Bond [772]

Answer:

S_{s}=300 m/s

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

Explanation:

In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.

The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.

For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

3 0
3 years ago
I need help please..its too complicated
Charra [1.4K]

Question: What was his initial velocity?

Answer: 3.62 m/s

3 0
2 years ago
A driver entering the outskirts of a city takes her foot off the accelerator so that the car slows down from 90 km/h to 50 km/h
Varvara68 [4.7K]

Answer:

Explanation:

a = (vf - vi) / t

a = (50 - 90) / 10.0

a = -4 km/h/s(1000 m/km / 3600 s/h)

a = - 1.11 m/s²

5 0
2 years ago
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