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:
not clear pic...but it's definitely not A)
The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
Read more about vertical weight here:
brainly.com/question/15244771
#SPJ1
I'd guess at valve B. more information about the interesting question would help.
Answer:
4.08 s
Explanation:
Let the passenger took "t" time to catch the train
so in this case the total distance moved by the train + 5 m = total distance moved by the passenger
so we will have
distance moved by train is given as
also the distance moved by passenger
so we will have
t = 4.08 s
