Using the appropriate approximations:
dx/L = mλ
d = slit separation
x = fringe spacing
L = distance between slits and screen
m = some integer, used to determine the distance from the central bright fringe to another bright fringe
We don't really need a value for m because we're calculating the distance between any pair of consecutive fringes. Let's just set m = 1
Given values:
d = 1.0mm
L = 2.0m
λ = 480nm
Substitute the terms in the equation with our given values and solve for x:
1.0*10⁻³*x/2 = 480*10⁻9
<h3>x = 0.96mm</h3>
Answer:
<h3>the momentum of the earth</h3>
Explanation:
<span>The best response would be that theories are not opinions. While laws and theories are similar, laws tend to be a collection of theories about the same phenomenon. Laws can be expressed as math formulae, but more often they're in the form of "if-then" statements. Laws and theories both have to be rigorously tested before they reach these classiffications.</span>
C
Because it’s B-A if u reared the question u will understand
Answer:
a) t=24s
b) number of oscillations= 11
Explanation:
In case of a damped simple harmonic oscillator the equation of motion is
m(d²x/dt²)+b(dx/dt)+kx=0
Therefore on solving the above differential equation we get,
x(t)=A₀
where A(t)=A₀
A₀ is the amplitude at t=0 and
is the angular frequency of damped SHM, which is given by,

Now coming to the problem,
Given: m=1.2 kg
k=9.8 N/m
b=210 g/s= 0.21 kg/s
A₀=13 cm
a) A(t)=A₀/8
⇒A₀
=A₀/8
⇒
applying logarithm on both sides
⇒
⇒
substituting the values

b) 

, where
is time period of damped SHM
⇒
let
be number of oscillations made
then, 
⇒