Angular width is 3 x 10^-3
Let D be the distance between source and screed d the distance between coherent source then for central diffraction maxima,
where λ is wavelength
Given:
λ = 450 nm = 450×10^−9m
d = 0.3x10^−3m, D = 1m
W = 2 x 450×10^−9/0.3x10^−3*1
To Find:
Angular width
Solution: The width of the central maxima is nothing but the difference between the positions of the first two minima. Hence we will use the expression for the position of minima and accordingly obtain the expression of the width of central maxima and secondary maxima
θ = W/D
θ = 2 x 450×10^−9/0.3x10^−3*1/1 = 3 x 10^-3
Hence, angular width is 3 x 10^-3
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Answer:
speed of eight ball speed after the collision is 3.27 m/s
Explanation:
given data
initially moving v1i = 3.4 m/s
final speed is v1f = 0.94 m/s
angle = θ w.r.t. original line of motion
solution
we assume elastic collision
so here using conservation of energy
initial kinetic energy = final kinetic energy .............1
before collision kinetic energy = 0.5 × m× (v1i)²
and
after collision kinetic energy = 0.5 × m× (v1f)² + 0.5 × m× (v2f)²
put in equation 1
0.5 × m× (v1i)² = 0.5 × m× (v1f)² + 0.5 × m× (v2f)²
(v2f)² = (v1i)² - (v1f)²
(v2f)² = 3.4² - 0.94²
(v2f)² = 10.68
taking the square root both
v2f = 3.27 m/s
speed of eight ball speed after the collision is 3.27 m/s
Answer:
they move towards the positive side... that's option 2
Answer:
7.13mL
Explanation:
P₁V₁T₁ = P₂V₂T₂
P₁ = 3atm , V₁ = 2.1 mL , T₁ = 273 + 4 =277K
P₂ = 0.95atm , V₂ = ? , T₂ = 273 + 25 =298K
V₂ = P₁V₁T₂ / P₂T₁
V₂ = (3atm)(2.1 mL )(298K) / (0.95atm)(277K)
V₂ = 7.13mL
