Answer:
a) A = 4.0 m
, b) w = 3.0 rad / s
, c) f = 0.477 Hz
, d) T = 20.94 s
Explanation:
The equation that describes the oscillatory motion is
x = A cos (wt + fi)
In the exercise we are told that the expression is
x = 4.0 cos (3.0 t + 0.10)
let's answer the different questions
a) the amplitude is
A = 4.0 m
b) the frequency or angular velocity
w = 3.0 rad / s
c) angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 3 / 2π
f = 0.477 Hz
d) the period
frequency and period are related
T = 1 / f
T = 1 / 0.477
T = 20.94 s
e) the phase constant
Ф = 0.10 rad
f) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф)
speed is maximum when sine is + -1
v = A w
v = 4 3
v = 12 m / s
g) the angular velocity is
w² = k / m
k = m w²
k = 1.2 3²
k = 10.8 N / m
h) the total energy of the oscillator is
Em = ½ k A²
Em = ½ 10.8 4²
Em = 43.2 J
i) the potential energy is
Ke = ½ k x²
for t = 0 x = 4 cos (0 + 0.1)
x = 3.98 m
j) kinetic energy
K = ½ m v²
for t = 00.1
²
v = A w sin 0.10
v = 4 3 sin 0.10
v = 1.98 m / s
Answer:
None of the above.
The correct answer would be momentum
Answer:
Work is the change in energy and power is the rate of doing work.
Answer:
660V
Explanation:
V=IR
V=11×60
=660V
hope this helps
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Answer:
0.0003 m = 0.3 mm
Explanation:
For constructive interference in the Young's experiment.
The position of the mth fringe from the central fringe is given by
y = L(mλ/d)
λ = wavelength = 720 nm = 720 × 10⁻⁹ m
L = distance between slits and screen respectively = 1.0 m
d = separation of slits = 0.68 mm = 0.68 × 10⁻³ m
m = 2
y = 1(2 × 720 × 10⁻⁹/(0.68 × 10⁻³) = 0.00212 m = 2.12 mm
For the 620 nm light,
y = 1(2 × 620 × 10⁻⁹/(0.68 × 10⁻³) = 0.00182 m = 1.82 mm
Distance apart = 2.12 - 1.82 = 0.3 mm = 0.0003 m
