Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
The spiral structure of the milky way can be explained by long lived quasi-static density waves<em>, </em><em>according to the lin-shu hypothesis. </em>Curiously, the waves of higher density gas and stars (seen as spiral arms) appear to remain static as stars move around the galaxy. This explained by assuming that the gravitational disturbances cause by the 'clumping' material in the arms does not affect the gravitational field of the galaxy as whole and is therefore negligible.
source: Astrophysicist
Answer:
14.2 m/s
Explanation:
Given data:
Speed of the stream, v₁ = 7.1 m/s
let the cross section area at initial point be A₁
now area at the second point, A₂ = (1/2)A₁ = 0.5A₁
now, from the continuity equation, we have
A₁v₁ = A₂v₂
where, v₂ is the velocity at the narrowed portion
thus, on substituting the values, we get
A₁ × 7.1 = 0.5A₁ × v₂
or
v₂ = 14.2 m/s
Answer:
500000000
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