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:
0.00091
Explanation:
(9x10^9) (2.6x10^-6) (1.4x10^-6) / 36
(9,000,000,000) (0.0000026) (0.0000014) /36
 |
 23,400(0.0000014) /36
 |
 0.03276 /36
 |
 0.00091
 
        
             
        
        
        
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