Explanation:
The water cycle is based on three parts;
1. Evaporation
2. Condensation
3. Participation
Condensation:
It is the process in which water vapor changes into liquid water or in other words, it is the transition from the gaseous state to liquid state.
Precipitation:
It is the process in which any liquid or frozen water such as snow that forms in the atmosphere and falls back to the Earth
Condensation depends on temperature and pressure whereas precipitation depends on the temperature and the concentration of the solution.
Answer:
0.012-m
Explanation:
∆L = α × Lo × (T-To)
α is the coefficient of linear expansion = 12 × 10-6 K-1
Lo = Initial length = 25-m
∆L = Change in length
(T-To) = 40 K
∆L = 12 × 10-6 × 25 × 40
∆L = 0.012-m
Answer:
False
Explanation:
A compass can be used to determine relative direction but not absolute direction.
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