Aluminium,metal,glass,and paper:
1 answer:
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Answer:
a) 9.72 mm
b) 4.86 mm
Explanation:
wave length of light λ is 580 nm = 580 \times 10⁻⁹ m
Width of slit d = 0.215\times 10⁻³ m
Distance of screen D = 1.8 m.
Width of one fringe =
Putting the values we get fringe width
=
=4.86 mm.
a) Width of central maxima = 2 times width of one fringe
= 2 times 4.86
=9.72 mm
b) width of each fringe except central fringe is same , no matter what the order is.Only brightness changes .
So width of either of the two first order bright fringe will be same and it will be
= 4.86 mm.
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 = / T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.