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
Air pollution level is higher in Location B, because poisonous fumes are produced when coal burns.
Answer:
Option (D) : The object slows down.
Answer:
The mass of Uranium present in a 1.2mg sample is 
Explanation:
The ration between Uranium mass and total sample mass is: 
For a sample of mass 1.2 mg, the amount of uranium is:
Answer:
a) The current is i = 1.2 A
b) The charge is Q = 17280 C
c) The energy is E = 43200 J
Explanation:
a) The current is given by the ohm's law wich is:
i = V/R = 3/2.5 = 1.2 A
b) Since the charge is steady we can use the following equation to find the charge amount in that time:
i = Q/t
Q = t*i
Where t is in seconds, so we have 4h * 3600 = 14400 s
Q = 1.2*14400 = 17280 C
c) The energy is the power delivered to the toy multiplied by the time:
P = 1.2*2.5 = 3 W
E = P*t = 3*14400 = 43200 J
