Answer:
The magnitude of the average induced emf is 90V
Explanation:
Given;
area of the square coil, A = 0.4 m²
number of turns, N = 15 turns
magnitude of the magnetic field, B = 0.75 T
time of change of magnetic field, t = 0.05 s
The magnitude of the average induced emf is given by;
E = -NAB/t
E = -(15 x 0.4 x 0.75) / 0.05
E = -90 V
|E| = 90 V
Therefore, the magnitude of the average induced emf is 90V
 
        
             
        
        
        
Answer:
i wish i knew your answer.
Explanation:
 
        
             
        
        
        
Answer:
a) k = 120 N / m
, b)    f = 0.851 Hz
, c)  v = 1,069 m / s
, d)  x = 0
, e)  a = 5.71 m / s²
, f)   x = 0.200 m
, g)  Em = 2.4 J
, h) v = -1.01 m / s
Explanation:
a) Hooke's law is
          F = k x
          k = F / x
           k = 24.0 / 0.200
           k = 120 N / m
b) the angular velocity of the simple harmonic movement is
         w = √ k / m
         w = √ (120 / 4.2)
         w = 5,345 rad / s
Angular velocity and frequency are related.
        w = 2π f
         f = w / 2π
         f = 5.345 / 2π
         f = 0.851 Hz
c) the equation that describes the movement is
         x = A cos (wt + Ф)
As the body is released without initial velocity, Ф = 0
         x = 0.2 cos wt
Speed is
        v = dx / dt
        v = -A w sin wt
The speed is maximum for sin wt = ±1
        v = A w
        v = 0.200 5.345
        v = 1,069 m / s
d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is
        x = A cos wt = 0
        x = 0
e) the acceleration is
        a = d²x / dt² = dv / dt
        a = - Aw² cos wt
The acceleration is maximum when cos wt = ± 1
        a = A w²
         a = 0.2   5.345
         a = 5.71 m / s²
f) the position for this acceleration is
        x = A cos wt
        x = A
        x = 0.200 m
g) Mechanical energy is
         Em = ½ k A²
         Em = ½ 120 0.2²
        Em = 2.4 J
h) the position is
          x = 1/3 A
Let's calculate the time to reach this point
          x = A cos wt
         1/3 A = A cos 5.345t
          t = 1 / w cos⁻¹(1/3)
The angles are in radians
 t = 1.23 / 5,345
 t = 0.2301 s
Speed is
 v = -A w sin wt
 v = -0.2 5.345 sin (5.345 0.2301)
 v = -1.01 m / s
i) acceleration
 a = -A w² sin wt
 a = - 0.2 5.345² cos (5.345 0.2301)
       a = -1.91 m / s²
 
        
             
        
        
        
I believe the answer is C: For objects at extremely fast speeds.
Hope this helps!