1) 
The capacitance of a parallel-plate capacitor is given by:

where
 is the vacuum permittivity
 is the vacuum permittivity
A is the area of each plate
d is the distance between the plates
Here, the radius of each plate is

so the area is

While the separation between the plates is

So the capacitance is

And now we can find the energy stored,which is given by:

2) 0.71 J/m^3
The magnitude of the electric field is given by

and the energy density of the electric field is given by

and using 
 , we find
, we find

 
        
             
        
        
        
Pushing, pulling is the answer
        
             
        
        
        
F = m*a 
30 N = (ma + mb) * a
30 = 5*a
a = 6 m/s ^2
F de B em A
30 - F de B,A = ma * a
30 - F de B em A = 3 * 6
30 - 18 = F de B em A
12 = F de B em A
Resposta: 6 m/s^2 e 12N
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Abç!
 
        
             
        
        
        
<u>Answer :</u>
(a) d = 0.25 m
(b) d = 0.5 m
<u>Explanation :</u>
It is given that,
Frequency of sound waves, f = 686 Hz
Speed of sound wave at  is, v = 343 m/s
 is, v = 343 m/s
(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.
 ........(1)
........(1)
Velocity of sound wave is given by :




Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.
(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.
 ( n = integers )
  ( n = integers )
Let n = 1 
So, 


Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.