Answer:
a) 0.0130 m
b') w' = =6.46*10^{-3] m
Explanation:
given data:
\lambda of light = 633 nm
width of siit a =0.360 mm
distance from screen = 3.75 m
a) the first minima is located at
=
with of central fringe = 2y_1 = 2*6.54 *10^{-3} = 0.0130 m
b)
width of the first bright fringe on either side of the central one = 
calculation for y_2
= 
w' = =6.46*10^{-3] m
Answer:
Newton's F=ma, which means the force (F) acting on an object is equal to the mass (m) of an object times its acceleration (a)
The lateral displacement is I don’t know tbh I think 16.8
The force needed to stretch the steel wire by 1% is 25,140 N.
The given parameters include;
- diameter of the steel, d = 4 mm
- the radius of the wire, r = 2mm = 0.002 m
- original length of the wire, L₁
- final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
- extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
- the Youngs modulus of steel, E = 200 Gpa
The area of the steel wire is calculated as follows;
The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;
Thus, the force needed to stretch the steel wire by 1% is 25,140 N.
Learn more here: brainly.com/question/21413915
