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:
A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.
Explanation:
Answer:
Electrons in atoms can act as our charge carrier, because every electron carries a negative charge. If we can free an electron from an atom and force it to move, we can create electricity.
Answer
1056
Explanation:
for example
A soil is to be excavated from a borrow pit which has a density of 1.75g/cc and water content of 12%. The G is 2.7 the soil is compacted to that water content of 18% and dry density of 1.65g/cc. for 1000 m3 of soil used in fill estimate
Quantity of soil to be excavated from pit in m3
Answer:
The final temperature in the vessel after the resistor has been operating for 30 min is 111.67°C
Explanation:
given information:
mass, m = 3 kg
initial temperature, T₁ = 40°C
current, I = 10 A
voltage, V = 50 V
time, t = 30 min = 1800 s
Heat for the system because of the resistance is
Q = V I t
where
V = voltage (V)
I = current (A)
t = time (s)
Q = heat transfer to the system (J)
so,
Q = V x I x t
= 50 x 10 x 1800
= 900000
= 9 x 10⁵ J
the heat transfer in the closed system is
Q = ΔU + W
where
U = internal energy
W = work done by the system
thus,
Q = ΔU + W
9 x 10⁵ = ΔU + 0, W = 0 because the tank is a well-insulated and rigid.
ΔU = 9 x 10⁵ J = 900 kJ
then, the energy change in the system is
ΔU = m c ΔT
ΔT = ΔU / m c, c = 4.186 J/g°C
= 900 / (3 x 4.186)
= 71.67°C
so,the final temperature (T₂)
ΔT = T₂ - T₁
T₂ = ΔT + T₁
= 71.67°C + 40°C
= 111.67°C
