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torisob [31]
3 years ago
6

Yung's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a sc

reen 1.10 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.2 mm from the center of the central bright fringe. What is the separation of the two slits?
Physics
1 answer:
Lady bird [3.3K]3 years ago
4 0

Answer:

1.082 mm

Explanation:

From the question, we can see that we were given The following

Wavelength of the atoms, λ = 502 nm = 502*10^-9 m

Radius of the screen away from the double slit, r = 1.1 m

We know that Y(20) = 10.2 mm = 10.2*10^-3 m

d = (20 * R * λ) / Y(20)

d = (20 * 1.1 * 502*10^-9)/10.2*10^-3

d = 1.1*10^-5 / 10.2*10^-3

d = 1.082 mm

Therefore, we can say that the distance of separation between the two slits is 1.082 mm

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(b) Moment of inertia through the edge of disk

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I=\frac{1}{2}MR^{2}\\  I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\  I=0.01 kg m^{2}

For (b) Moment of inertia through the edge of disk

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3 years ago
Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 x 10^4 Pa. If
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Answer:

p2 = 9.8×10^4 Pa

Explanation:

Total pressure is constant and PT = P = 1/2×ρ×v^2  

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from continuity we have ρ×A1×v1 = ρ×A2×v2  

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v2 = (v1)/4  

then:

p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2  

p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)  

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