Answer:
It is not connected by any way
Answer:
(h₁-h₂) = 2.30 10² m
Explanation:
The pressure depends on the height with the formula
P = P_atm + rho g h
Let's apply this expression for the building
P₁ = P_atm + rho_air g h₁
P₂ = P_atm + rho_air g h₂
Subtract
P₁ - P₂ = roh_air g (h₁ –h₂)
The measured pressure is in mm Hg to take this unit to units of pressure must be multiplied by the density of mercury and the acceleration of gravity
P₁- P₂ = rho_Hg g (h₁-h₂) _Hg
rho_Hg g (h₁-h₂) _Hg = roh_air g (h₁ –h₂)
(h₁ –h₂) = rho_Hg / rho_air (h₁-h₂) _ Hg
Let's calculate
(h₁-h₂) = 13600 / 1.18 (695-675)
(h₁-h₂) = 2.30 10⁵ mm
Let's reduce to meter
(h₁-h₂) = 2.30 10⁵ mm (1 m / 10³ mm)
(h₁-h₂) = 2.30 10² m
Quasi frequency = 4√6
Quasi period = π√6/12
t ≈ 0.4045
<u>Explanation:</u>
Given:
Mass, m = 20g
τ = 400 dyn.s/cm
k = 3920
u(0) = 2
u'(0) = 0
General differential equation:
mu" + τu' + ku = 0
Replacing the variables with the known value:
20u" + 400u' + 3920u = 0
Divide each side by 20
u" + 20u' + 196u = 0
Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.
r² + 20r + 196 = 0
Determining the roots:

r = -10 ± 4√6i
The general solution for two complex roots are:
y = c₁ eᵃt cosbt + c₂ eᵃt sinbt
with a the real part of the roots and b be the imaginary part of the roots.
Since, a = -10 and b = 4√6
u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t
u(0) = 2
u'(0) = 0
(b)
Quasi frequency:
μ = 

(c)
Quasi period:
T = 2π / μ

(d)
|u(t)| < 0.05 cm
u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05
solving for t:
τ = t ≈ 0.4045
1.It gives less load on the shaft.
2.It transmits more power than belts.
3.It may be used for both long as well as short distance.
If the amount of electrical energy is 50 Joules before the conversions, then it would be 50 Joules after the conversion.
According to law of conservation of energy, we cannot create or destroy energy so it remains constant
Hope this helps!