Answer:
The width of the strand of hair is 1.96 10⁻⁵ m
Explanation:
For this diffraction problem they tell us that it is equivalent to the diffraction of a single slit, which is explained by the equation
<h3> a sin θ =± m λ
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Where the different temrs are: “a” the width of the hair, λ the wavelength, θ the angle from the center, m the order of diffraction, which is the number of bright rings (constructive diffraction)
We can see that the diffraction angle is missing, but we can find it by trigonometry, where L is the distance of the strand of hair to the observation screen and "y" is the perpendicular distance to the first minimum of intensity
L = 1.25 m 100 cm/1m = 125 cm
y = 5.06 cm
Tan θ = y/L
Tan θ = 5.06/125
θ = tan⁻¹ ( 0.0405)
θ = 2.32º
With this data we can continue analyzing the problem, they indicate that they measure the distance to the first dark strip, thus m = 1
a = m λ / sin θ
a = 1 633 10⁻⁹ 1.25/sin 2.3
a = 1.96 10⁻⁵ m
a = 0.0196 mm
The width of the strand of hair is 1.96 10⁻⁵ m
Answer:
Partial Pressure of F₂ = 1.30 atm
Partial pressure of Cl₂ = 0.70 atm
Explanation:
Partial pressure for gases are given by Daltons law.
Total pressure of a gas mixture = sum of the partial pressures of individual gases
Pt = P(f₂) + P(cl₂)
Partial pressure = mole fraction × total pressure
Let the mass of each gas present be m
Number of moles of F₂ = m/38 (molar mass of fluorine = 38 g/Lol
Number of moles of Cl₂ = m/71 (molar mass of Cl₂)
Mole fraction of F₂ = (m/38)/((m/38) + (m/71)) = 0.65
Mole fraction of Cl₂ = (m/71)/((m/38) + (m/71)) = 0.35 or just 1 - 0.65 = 0.35
Partial Pressure of F₂ = 0.65 × 2 = 1.30 atm
Partial pressure of Cl₂ = 0.35 × 2 = 0.70 atm
If n=1 you divide by 2
If n=2 you divide by 4
If n=3 you divide by 8
so any n you divide by 2 to the power n
When the surface of the comb rubs on your hair, the comb is electrically charged. When the comb comes close to the paper, the charge on the comb causes charge separation on the paper bits. Since paper is neutral, positive and negative charges are equivalent. The charge on the comb charges the area of the bit of paper nearest the comb to the opposite. Thus, the bits of paper become attracted to the comb.
