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Anastasy [175]
3 years ago
15

I need help on this I'm not the best at sicence

Physics
1 answer:
AveGali [126]3 years ago
4 0
Science
ask me in comments

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You might be interested in
In a Young's double-slit experiment, a set of parallel slits with a separation of 0.102 mm is illuminated by light having a wave
guajiro [1.7K]

Answer:

Rounded to three significant figures:

(a) 2 \times 575\; \rm nm = 1150\; \rm nm = 1.15\times 10^{-6}\; \rm m.

(b) \displaystyle \left(1 + \frac{1}{2}\right) \times (575\;\rm nm) \approx 863\; \rm nm = 8.63\times 10^{-7}\; \rm m.

Explanation:

Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.  

Let k denote a natural number (k \in \left\lbrace0,\, 1,\, 2,\, \dots\right\rbrace.) In a double-split experiment of a monochromatic light:

  • A maximum (a bright fringe) is produced when light from the two slits arrive while they were in-phase. That happens when the path difference is an integer multiple of wavelength. That is: \text{Path difference} = k\, \lambda.
  • Similarly, a minimum (a dark fringe) is produced when light from the two slits arrive out of phase by exactly one-half of the cycle. For example, The first wave would be at peak while the second would be at a crest when they arrive at the screen. That happens when the path difference is an integer multiple of wavelength plus one-half of the wavelength: \displaystyle \text{Path difference} = \left(k + \frac{1}{2}\right)\cdot \lambda.
<h3 /><h3>Maxima</h3>

The path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where k = 0 in \text{Path difference} = 0.

The path difference increases while moving on the screen away from the center. The first order maximum is at k = 1 where \text{Path difference} = \lambda.

Similarly, the second order maximum is at k = 2 where \text{Path difference} = 2\, \lambda. For the light in this question, at the second order maximum: \text{Path difference} = 2\, \lambda = 2 \times 575\; \rm nm = 1.15\times 10^{-6}\; \rm m.

  • Central maximum: k = 0, such that \text{Path difference} = 0.
  • First maximum: k = 1, such that \text{Path difference} = \lambda.
  • Second maximum: k = 2, such that \text{Path difference} = 2\, \lambda.

<h3>Minima</h3>

The dark fringe closest to the center of the screen is the first minimum. \displaystyle \text{Path difference} = \left(0 + \frac{1}{2}\right)\cdot \lambda = \frac{1}{2}\, \lambda at that point.

Add one wavelength to that path difference gives another dark fringe- the second minimum. \displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda at that point.

  • First minimum: k =0, such that \displaystyle \text{Path difference} = \frac{1}{2}\, \lambda.
  • Second minimum: k =1, such that \displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda.

For the light in this question, at the second order minimum: \displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda =  \left(1 + \frac{1}{2}\right)\times (575\; \rm nm) \approx 8.63\times 10^{-7}\; \rm m.

6 0
3 years ago
During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±
Alla [95]

Answer:

P = 1 (14,045 ± 0.03 )  k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

            Δ (Pₓ / Py) =?

 Let's start by finding the momentum of each vehicle

car X

        Pₓ = m vₓ

        Pₓ = 2.34 2.5

        Pₓ = 5.85 kg m

car Y

        Py = 2,561 3.2

        Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

          ΔPₓ = m Δv + v Δm

          ΔPₓ = 2.34 0.01 + 2.561 0.01

          ΔPₓ = 0.05 kg m

         ΔP_{y} = m Δv + v Δm

         ΔP_{y} = 2,561 0.01+ 3.2 0.001

         ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

          P = Pₓ / P_{y}

          ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

          ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

          ΔP = 0.006 + 0.0026

          ΔP = 0.009 kg m

The result is

           P = 14,045 ± 0.039 = (14,045 ± 0.03 )  k gm/s

7 0
4 years ago
A television of mass 8 kg sits on a table. The coefficient of static friction
grandymaker [24]

The television has weight (8 kg) <em>g</em> = 78.4 N, and the magnitude of the normal force between the table and television would be the same, 78.4 N. This mean the maximum magnitude of static friction between the table and television is

0.48 (78.4 N) ≈ 37.6 N ≈ 38 N

and this is the minimum required force needed to get the television to slide.

5 0
4 years ago
The first model of the atom was developed through what?
Dennis_Churaev [7]
Democritus Introduces the Atom
He thought that a point would be reached where matter could not be cut into still smaller pieces. He called these "uncuttable" pieces atomos. This is where the modern term atom comes from. Democritus first introduced the idea of the atom almost 2500 years ago.
5 0
3 years ago
Read 2 more answers
Is it possible for a gas contained in a chamber to maintain a constant temperature while heat is being added to the gas? Explain
Evgesh-ka [11]

Answer:Yes

Explanation:

Yes it is possible for a gas contained in a chamber to maintain a constant temperature while heat is being added to the gas.A process in which temperature of the gas remains constant is called Isothermal Process.For an ideal internal energy is a function of temperature therefore internal energy remains constant while all the heat added is converted to do the work done by the system.

6 0
4 years ago
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