Answer:
This is the answer of your question.
What type of energy transfer occurs through light or electromagnetic waves? A) conduction B) radiation C) induction
2 answers:
You might be interested in
The name of this landmas is known as <em></em>Pangaea, was a supercontinent that existed during the late Paleozoic and early Mesozoic eras. It formed approximately 300 million years ago and began to break apart after about 100 million years.
Theres an image of how this supercontinet looked
Theres an image of how this supercontinet looked
Answer with Explanation:
From work energy theorem we have
Now since it is given that the work done on the object by the force is positive hence we conclude that the term on the right hand of the above relation is positive hence
Part a)
Hence we conclude that the mechanical energy of the particle will increase.
Part b) Since the mechanical energy of the particle is the sum of it's kinetic and potential energies, we can write
Now since the sum of the 2 energies ( Kinetic and potential ) is positive we cannot be conclusive of the individual values since 2 cases may arise:
1) Kinetic energy increases while as potential energy decreases.
2) Potential energy increases while as kinetic energy decreases.
Hence these two cases are possible and we cannot find using only the given information which case will hold
Hence no conclusion can be formed regarding the individual energies.
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
