The classical motion for an oscillator that starts from rest at location x₀ is
x(t) = x₀ cos(ωt)
The probability that the particle is at a particular x at a particular time t
is given by ρ(x, t) = δ(x − x(t)), and we can perform the temporal average
to get the spatial density. Our natural time scale for the averaging is a half
cycle, take t = 0 → π/
ω
Thus,
ρ = 
Limit is 0 to π/ω
We perform the change of variables to allow access to the δ, let y = x₀ cos(ωt) so that
ρ(x) = 
Limit is x₀ to -x₀

Limit is -x₀ to x₀

This has
as expected. Here the limit is -x₀ to x₀
The expectation value is 0 when the ρ(x) is symmetric, x ρ(x) is asymmetric and the limits of integration are asymmetric.
Answer:
<u>the thickness required would be 12 inch HMA and granular base layer of 6 inches</u>
Explanation:
structural number = 4.5
stone base course material coefficient = 0.13
hma material layer coefficient = 0.40
drainage coefficient = 0.90
we will use layered analysis procedure to get thickness
D1 >= sN1/a1
when we cross multiply,
sN1 = a1D1 >=sN1
D2 >= -sN2-sN1/a2m2
sN2* + sN1* >= sN2
D3 >= sN3-(sN1*+sN2*)/a2m2
where a1,a2,a3 = layer coefficient
d1 d2 d3 = actual thickness
m2,m3 = coefficient of base
a1 = 0.4
a1 = 0.13
sN = 4.5
m2 = 0.9
D1 >= sN1/a1 = 4.5/0.4
= 11.25
thickness of surface = 12 inches
a1D1 = 0.4x12 = 4.8
we have value of sN2 = 5.5
(5.5 -4.8)/(0.13*0.9)
= 0.7/0.117
= 5.9829 inches
approximately 6 inches
so the pavement will have 12inch HMA surface and 6 inches granular base layer.
Answer: hello attached below is the properly written chain reaction to your question
answer :
d[NO] / dt = ![k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] + K_{3f}[N][OH]](https://tex.z-dn.net/?f=k_%7B1f%7D%20%5BO%5D%20%5BN_%7B2%7D%5D%20%2B%20K_%7B2f%7D%20%20%5BN%5D%5BO_%7B2%7D%20%5D%20%2B%20K_%7B3f%7D%5BN%5D%5BOH%5D)
d[N] / dt = ![k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH]](https://tex.z-dn.net/?f=k_%7B1f%7D%20%5BO%5D%20%5BN_%7B2%7D%5D%20%2B%20K_%7B2f%7D%20%20%5BN%5D%5BO_%7B2%7D%20%5D%20-%20K_%7B3f%7D%5BN%5D%5BOH%5D)
Explanation:
<u>write out expressions for d[NO] / dt and d[N] / dt</u>
Given :
properly written chain reaction ( attached below)
Expression for d[NO] / dt can be written as
![k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] + K_{3f}[N][OH]](https://tex.z-dn.net/?f=k_%7B1f%7D%20%5BO%5D%20%5BN_%7B2%7D%5D%20%2B%20K_%7B2f%7D%20%20%5BN%5D%5BO_%7B2%7D%20%5D%20%2B%20K_%7B3f%7D%5BN%5D%5BOH%5D)
Expression for d[N] / dt can be written as
![k_{1f} [O] [N_{2}] + K_{2f} [N][O_{2} ] - K_{3f}[N][OH]](https://tex.z-dn.net/?f=k_%7B1f%7D%20%5BO%5D%20%5BN_%7B2%7D%5D%20%2B%20K_%7B2f%7D%20%20%5BN%5D%5BO_%7B2%7D%20%5D%20-%20K_%7B3f%7D%5BN%5D%5BOH%5D)
Answer:
Ignition switch, neutral safety switch, the starter solenoid, starter motor, and finally the batteries.
Explanation: