Answer:
Mechanical resonance frequency is the frequency of a system to react sharply when the frequency of oscillation is equal to its resonant frequency (natural frequency).
The physical dimension of the silicon is 10kg
Explanation:
Using the formular, Force, F = 1/2π√k/m
At resonance, spring constant, k = mw² ( where w = 2πf), when spring constant, k = centripetal force ( F = mw²r).
Hence, F = 1/2π√mw²/m = f ( f = frequency)
∴ f = F = mg, taking g = 9.8 m/s²
100 Hz = 9.8 m/s² X m
m = 100/9.8 = 10.2kg
Answer:
a)W=12.62 kJ/mol
b)W=12.59 kJ/mol
Explanation:
At T = 100 °C the second and third virial coefficients are
B = -242.5 cm^3 mol^-1
C = 25200 cm^6 mo1^-2
Now according isothermal work of one mole methyl gas is
W=-
a=
b=
from virial equation

And

a=
b=
Now calculate V1 and V2 at given condition

Substitute given values
= 1 x 10^5 , T = 373.15 and given values of coefficients we get

Solve for V1 by iterative or alternative cubic equation solver we get

Similarly solve for state 2 at P2 = 50 bar we get

Now

a=241.33
b=30780
After performing integration we get work done on the system is
W=12.62 kJ/mol
(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get
dV=RT(-1/p^2+0+C')dP
Hence work done on the system is

a=
b=
by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work
W=12.59 kJ/mol
The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series
Answer:
It involves the active streamlining of a business's supply-side activities to maximize customer value and gain a competitive advantage in the marketplace
Explanation:
Supply chain management is the management of the flow of goods and services and includes all processes that transform raw materials into final products.
Answer:
(a) Flow rate of vehicles = No of vehicles per mile * Speed
=No of cars per mile * Speed +No of trucks per mile * Speed
= 0.75*50*60 + 0.25*50*40
=2750 vehicles / hour
(b) Let Density of vehicles on grade = x
Density on flat * Speed =Density on grade * Speed
So,( 0.75*50) * 60 + (0.25*50) * 40 = (0.75* x) * 55 + (0.25* x) * 25
So, x= 57.89
So, Density is around 58 Vehicles per Mile.
(c) Percentage of truck by aerial photo = 25%
(d)Percentage of truck bystationary observer on the grade= 25*30/60 * 25/55 =22.73 %