The force of gravity is the
force with which massively large objects such as the earth attracts another
object towards itself. All objects of the earth exert a gravity that is
directed towards the center of the earth. Therefore, the force of gravity of
the earth is equal to the mass of the object times acceleration due to gravity and also equal to the weight of the object.
F = ma
since F = W
W = ma
Answer
When the light goes from one medium to another medium with higher refractive index the velocity of the ray decreases, wavelength of the ray also decreases.
But the frequency of the ray when it enters the medium of higher refractive index remain same.
So, we can conclude that speed of ray and wavelength decrease but frequency remain unchanged.
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:
Explanation:
v² = u² + 2as
s = (v² - u²) / 2a
s = (29.88² - 6.73²) / (2(5.22))
s = 81.1802203065...
s = 81.18 m
Answer: Amplitude
Explanation: The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates (i.e the amplitude of vibration).
Loudness is measured in units called decibels and amplitude in metre