Answer:
46 dgres is the answer
Explanation:
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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
Answer:
Physical property
Explanation:
Physical characteristic of matter is not associated with the changes in its chemical composition.Some of these properties are color and density that can be observed without altering the physical state of the matter.Other physical properties such as freezing and melting are observed when a change in physical state of matter. When there is not change in the identities of substances contained in the matter then that's a physical change property.
Answer:
R = V^2 sin 2 θ / g range formula
R is a maximum for θ = 45 and R = V^2 g
g = V^2 / R = 3^2 / 15 = .6 m/s^2
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