im sorry but i dont know, good luck at finding someone else who does.
 
        
             
        
        
        
Answer:

Explanation:
When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

Where  is the length of the string and
 is the length of the string and  the velocity of propagation. Use this expression to find the value of
 the velocity of propagation. Use this expression to find the value of  .
.

The velocity of propagation is given by the expression:

Where  is the desirable variable of the problem, the linear mass density, and
 is the desirable variable of the problem, the linear mass density, and  is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:
 is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

With the value of the tension and the velocity you can find the mass density:


 
        
             
        
        
        
Answer:1.5×10 to the power of 17(unit-Hertz/H)
Explanation:V=F×Wavelength
F=V/Wavelength=3×10 to power/2×10 to power of -9=1.5×10 to power of 17
 
        
             
        
        
        
Answer:
The torque about his shoulder is 34.3Nm. 
The solution approach assumes that the weight of the boy's arm acts at the center of the boy's arm length 35cm from the shoulder.
Explanation:
The solution to the problem can be found in the attachment below.
 
        
             
        
        
        
Answer
 Pressure, P = 1 atm
 air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
    Pressure at sea level = 1 atm = 101300 Pa
    we know
    P = ρ g h
    
    
           h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
   at x = 0  air density = 0
   at x= h   ρ_l = ρ_sl
  assuming density is zero at x - distance
  
now, Pressure at depth x
 
 
integrating both side
 

  now,


   h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.