Answer:
The potential energy can be given as
E = mgh. m is mass, g = acceleration due to gravity = 9.8m/s, h is the heigh, given as 100.0m
E = m x 9.8 x 100 = (980m)J
E = (980m)/10^9GJ = (0.000000980m)GJ to 3 significant figures
Explanation:
Hydroelectric dams exploit storage of gravitational potential energy. A mass, m, raised a height, h against gravity, g = 9.8 m/s², is given a potential energy E = mgh. The result will be in Joules if the input is expressed in meters, kilograms, and seconds (MKS, or SI units).
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
Refer to the diagram shown below.
h = height of the girl above water when she lets go of the rope.
The launch velocity is 22.5 m/s at 35° to the horizontal. Therefore the vertical component of the velocity is
v = 22.5 sin(35°) = 12.9055 m/s.
The time of flight is t = 1.10 s before the girl hits the surface of the water at a height of -h.
Therefore
-h = (12.9055 m/s)*(1.10 s) - (1/2)*(9.8 m/s²)*(1.10 s)²
-h = 8.267 m
= 8.3 m (nearest tenth)
Answer:
When the girl let go of the rope, she was about 8.3 m above the surface of the water.
METAL, BRICKS, WOOD,AND ECT